Mixing
Solving or bounding the real part of the integral $int_0^{2 pi i m} frac{e^{-t}}{t-a} dt$
$begingroup$
I would be interested in finding a closed form or, at least, bounding (in terms of $m$ as it becomes larger) the real part of the following itnegral:
$$f(m,a):=int_0^{2 pi i m} frac{e^{-t}}{t-a} dt$$
where $m in mathbb{N}, m>>0$ and $a in mathbb{C}, mathfrak{Re}( a )>>0$.
For this purpose, I have tried three different methods.
Attempt 1: Firstly, I just tried to express the integral as a sum of integrals and made a change of variables, so that
$$f(m,a)= i sum_{n=0}^{m-1} int_{2 pi n}^{2 pi (n+1)} frac{e^{-it}}{it-a} dt$$
Then, I compared those integrals whith the following:
$$int_{gamma} frac{1}{log{z}-a} dz $$
Where the path of integration is the unit circle clockwise. Each integral from sum of integrals correspond to different branches of the complex logarithm. To analyse this last one, I used a keyhole contour, taking the branch cut of the logarithm at the positive real axis. I obtained, for example, for the first summand (corresponding to the first branch of the logarithm),
$$int_{0}^{2 pi} frac{e^{-it}}{it-a} dt = iRint_0^{2 pi} frac{e^{it}}{2 pi i - log{R} -i t -a}dt - int_1^R frac{2 i pi}{(a+log{t})(a+log{t}-2 i pi)}dt$$
For $R in mathbb{R}, R > |a|$. My problem is that this family of integrals seems even more difficult to bound (even letting $R to infty$), so I have not gained anything.
Attempt 2: We could multiply and divide by $e^a$, so that we get:
$$f(m,a)=e^{-a} int_0^{2 pi i m} frac{e^{-(t-a)}}{t-a} dt$$
From here, we could relate it to the Exponential Integral. However, the bounds that can be obtained this way seem to be not accurate at all for big $m$ and $mathfrak{Re} a$, and it is difficult to extract the real part from them.
Attempt 3: We could separe the real and imaginary part of the original integral:
$$f(m,a)= -i int_0^{2 pi m} frac{(cos{t} + i sin{t})(ileft (t + mathfrak{Im}(a) right) + mathfrak{Re}(a))}{mathfrak{Re}(a)^2+left (mathfrak{Im}(a) + t right) ^2} dt$$
$$mathfrak{Re}(f(m,a))= int_0^{2 pi m} frac{left (mathfrak{Im}(a) + t right) cos{t} + mathfrak{Re}(a) sin{t}}{mathfrak{Re}(a)^2+left (mathfrak{Im}(a) + t right) ^2} dt$$
However, I do not know how to solve or bound this last integral neither.
Attempt 4: Using the Fourier Transform method mentiones in the answer, we can work around with the integrals, but would end up with Exponential Integral (attempt 2).
Any help will be welcomed.
Thank you.
integration definite-integrals contour-integration fourier-transform
asked Jan 29 at 10:51
user3141592user3141592
859623
$endgroup$
add a comment |
$begingroup$
I would be interested in finding a closed form or, at least, bounding (in terms of $m$ as it becomes larger) the real part of the following itnegral:
$$f(m,a):=int_0^{2 pi i m} frac{e^{-t}}{t-a} dt$$
where $m in mathbb{N}, m>>0$ and $a in mathbb{C}, mathfrak{Re}( a )>>0$.
For this purpose, I have tried three different methods.
Attempt 1: Firstly, I just tried to express the integral as a sum of integrals and made a change of variables, so that
$$f(m,a)= i sum_{n=0}^{m-1} int_{2 pi n}^{2 pi (n+1)} frac{e^{-it}}{it-a} dt$$
Then, I compared those integrals whith the following:
$$int_{gamma} frac{1}{log{z}-a} dz $$
Where the path of integration is the unit circle clockwise. Each integral from sum of integrals correspond to different branches of the complex logarithm. To analyse this last one, I used a keyhole contour, taking the branch cut of the logarithm at the positive real axis. I obtained, for example, for the first summand (corresponding to the first branch of the logarithm),
$$int_{0}^{2 pi} frac{e^{-it}}{it-a} dt = iRint_0^{2 pi} frac{e^{it}}{2 pi i - log{R} -i t -a}dt - int_1^R frac{2 i pi}{(a+log{t})(a+log{t}-2 i pi)}dt$$
For $R in mathbb{R}, R > |a|$. My problem is that this family of integrals seems even more difficult to bound (even letting $R to infty$), so I have not gained anything.
Attempt 2: We could multiply and divide by $e^a$, so that we get:
$$f(m,a)=e^{-a} int_0^{2 pi i m} frac{e^{-(t-a)}}{t-a} dt$$
From here, we could relate it to the Exponential Integral. However, the bounds that can be obtained this way seem to be not accurate at all for big $m$ and $mathfrak{Re} a$, and it is difficult to extract the real part from them.
Attempt 3: We could separe the real and imaginary part of the original integral:
$$f(m,a)= -i int_0^{2 pi m} frac{(cos{t} + i sin{t})(ileft (t + mathfrak{Im}(a) right) + mathfrak{Re}(a))}{mathfrak{Re}(a)^2+left (mathfrak{Im}(a) + t right) ^2} dt$$
$$mathfrak{Re}(f(m,a))= int_0^{2 pi m} frac{left (mathfrak{Im}(a) + t right) cos{t} + mathfrak{Re}(a) sin{t}}{mathfrak{Re}(a)^2+left (mathfrak{Im}(a) + t right) ^2} dt$$
However, I do not know how to solve or bound this last integral neither.
Attempt 4: Using the Fourier Transform method mentiones in the answer, we can work around with the integrals, but would end up with Exponential Integral (attempt 2).
Any help will be welcomed.
Thank you.
integration definite-integrals contour-integration fourier-transform
asked Jan 29 at 10:51
user3141592user3141592
859623
$endgroup$
$begingroup$
What do you mean by bounding? Something like $$left| int_0^{2 pi i m} frac{e^{-t}}{t-a} , {rm d}t right| leq frac{2pi m}{|a|^2} left( 1+frac{2}{|a|} right) , ?$$
$endgroup$
– Diger
Jan 30 at 0:00
$begingroup$
@Diger I do expect a better bound, since it seems clear that, for fixed $a$, the limit as $mathfrak{Re} (m) to infty$ of the real part of the integral converges (or is of the form $C + O ( frac{1}{m} ) $ for a constant $C$). I would like to upper and lower bound this result as $m$ becomes larger
$endgroup$
– user3141592
Jan 30 at 9:35
add a comment |
$begingroup$
I would be interested in finding a closed form or, at least, bounding (in terms of $m$ as it becomes larger) the real part of the following itnegral:
$$f(m,a):=int_0^{2 pi i m} frac{e^{-t}}{t-a} dt$$
where $m in mathbb{N}, m>>0$ and $a in mathbb{C}, mathfrak{Re}( a )>>0$.
For this purpose, I have tried three different methods.
Attempt 1: Firstly, I just tried to express the integral as a sum of integrals and made a change of variables, so that
$$f(m,a)= i sum_{n=0}^{m-1} int_{2 pi n}^{2 pi (n+1)} frac{e^{-it}}{it-a} dt$$
Then, I compared those integrals whith the following:
$$int_{gamma} frac{1}{log{z}-a} dz $$
Where the path of integration is the unit circle clockwise. Each integral from sum of integrals correspond to different branches of the complex logarithm. To analyse this last one, I used a keyhole contour, taking the branch cut of the logarithm at the positive real axis. I obtained, for example, for the first summand (corresponding to the first branch of the logarithm),
$$int_{0}^{2 pi} frac{e^{-it}}{it-a} dt = iRint_0^{2 pi} frac{e^{it}}{2 pi i - log{R} -i t -a}dt - int_1^R frac{2 i pi}{(a+log{t})(a+log{t}-2 i pi)}dt$$
For $R in mathbb{R}, R > |a|$. My problem is that this family of integrals seems even more difficult to bound (even letting $R to infty$), so I have not gained anything.
Attempt 2: We could multiply and divide by $e^a$, so that we get:
$$f(m,a)=e^{-a} int_0^{2 pi i m} frac{e^{-(t-a)}}{t-a} dt$$
From here, we could relate it to the Exponential Integral. However, the bounds that can be obtained this way seem to be not accurate at all for big $m$ and $mathfrak{Re} a$, and it is difficult to extract the real part from them.
Attempt 3: We could separe the real and imaginary part of the original integral:
$$f(m,a)= -i int_0^{2 pi m} frac{(cos{t} + i sin{t})(ileft (t + mathfrak{Im}(a) right) + mathfrak{Re}(a))}{mathfrak{Re}(a)^2+left (mathfrak{Im}(a) + t right) ^2} dt$$
$$mathfrak{Re}(f(m,a))= int_0^{2 pi m} frac{left (mathfrak{Im}(a) + t right) cos{t} + mathfrak{Re}(a) sin{t}}{mathfrak{Re}(a)^2+left (mathfrak{Im}(a) + t right) ^2} dt$$
However, I do not know how to solve or bound this last integral neither.
Attempt 4: Using the Fourier Transform method mentiones in the answer, we can work around with the integrals, but would end up with Exponential Integral (attempt 2).
Any help will be welcomed.
Thank you.
integration definite-integrals contour-integration fourier-transform
asked Jan 29 at 10:51
user3141592user3141592
859623
$endgroup$
I would be interested in finding a closed form or, at least, bounding (in terms of $m$ as it becomes larger) the real part of the following itnegral:
$$f(m,a):=int_0^{2 pi i m} frac{e^{-t}}{t-a} dt$$
where $m in mathbb{N}, m>>0$ and $a in mathbb{C}, mathfrak{Re}( a )>>0$.
For this purpose, I have tried three different methods.
Attempt 1: Firstly, I just tried to express the integral as a sum of integrals and made a change of variables, so that
$$f(m,a)= i sum_{n=0}^{m-1} int_{2 pi n}^{2 pi (n+1)} frac{e^{-it}}{it-a} dt$$
Then, I compared those integrals whith the following:
$$int_{gamma} frac{1}{log{z}-a} dz $$
Where the path of integration is the unit circle clockwise. Each integral from sum of integrals correspond to different branches of the complex logarithm. To analyse this last one, I used a keyhole contour, taking the branch cut of the logarithm at the positive real axis. I obtained, for example, for the first summand (corresponding to the first branch of the logarithm),
$$int_{0}^{2 pi} frac{e^{-it}}{it-a} dt = iRint_0^{2 pi} frac{e^{it}}{2 pi i - log{R} -i t -a}dt - int_1^R frac{2 i pi}{(a+log{t})(a+log{t}-2 i pi)}dt$$
For $R in mathbb{R}, R > |a|$. My problem is that this family of integrals seems even more difficult to bound (even letting $R to infty$), so I have not gained anything.
Attempt 2: We could multiply and divide by $e^a$, so that we get:
$$f(m,a)=e^{-a} int_0^{2 pi i m} frac{e^{-(t-a)}}{t-a} dt$$
From here, we could relate it to the Exponential Integral. However, the bounds that can be obtained this way seem to be not accurate at all for big $m$ and $mathfrak{Re} a$, and it is difficult to extract the real part from them.
Attempt 3: We could separe the real and imaginary part of the original integral:
$$f(m,a)= -i int_0^{2 pi m} frac{(cos{t} + i sin{t})(ileft (t + mathfrak{Im}(a) right) + mathfrak{Re}(a))}{mathfrak{Re}(a)^2+left (mathfrak{Im}(a) + t right) ^2} dt$$
$$mathfrak{Re}(f(m,a))= int_0^{2 pi m} frac{left (mathfrak{Im}(a) + t right) cos{t} + mathfrak{Re}(a) sin{t}}{mathfrak{Re}(a)^2+left (mathfrak{Im}(a) + t right) ^2} dt$$
However, I do not know how to solve or bound this last integral neither.
Attempt 4: Using the Fourier Transform method mentiones in the answer, we can work around with the integrals, but would end up with Exponential Integral (attempt 2).
Any help will be welcomed.
Thank you.
integration definite-integrals contour-integration fourier-transform
integration definite-integrals contour-integration fourier-transform
asked Jan 29 at 10:51
user3141592user3141592
859623
asked Jan 29 at 10:51
user3141592user3141592
859623
edited Feb 21 at 13:22
user3141592
asked Jan 29 at 10:51
user3141592user3141592
859623
asked Jan 29 at 10:51
user3141592user3141592
859623
asked Jan 29 at 10:51
user3141592user3141592
859623
859623
$begingroup$
What do you mean by bounding? Something like $$left| int_0^{2 pi i m} frac{e^{-t}}{t-a} , {rm d}t right| leq frac{2pi m}{|a|^2} left( 1+frac{2}{|a|} right) , ?$$
$endgroup$
– Diger
Jan 30 at 0:00
$begingroup$
@Diger I do expect a better bound, since it seems clear that, for fixed $a$, the limit as $mathfrak{Re} (m) to infty$ of the real part of the integral converges (or is of the form $C + O ( frac{1}{m} ) $ for a constant $C$). I would like to upper and lower bound this result as $m$ becomes larger
$endgroup$
– user3141592
Jan 30 at 9:35
add a comment |
$begingroup$
What do you mean by bounding? Something like $$left| int_0^{2 pi i m} frac{e^{-t}}{t-a} , {rm d}t right| leq frac{2pi m}{|a|^2} left( 1+frac{2}{|a|} right) , ?$$
$endgroup$
– Diger
Jan 30 at 0:00
$begingroup$
@Diger I do expect a better bound, since it seems clear that, for fixed $a$, the limit as $mathfrak{Re} (m) to infty$ of the real part of the integral converges (or is of the form $C + O ( frac{1}{m} ) $ for a constant $C$). I would like to upper and lower bound this result as $m$ becomes larger
$endgroup$
– user3141592
Jan 30 at 9:35
$begingroup$
What do you mean by bounding? Something like $$left| int_0^{2 pi i m} frac{e^{-t}}{t-a} , {rm d}t right| leq frac{2pi m}{|a|^2} left( 1+frac{2}{|a|} right) , ?$$
$endgroup$
– Diger
Jan 30 at 0:00
$begingroup$
What do you mean by bounding? Something like $$left| int_0^{2 pi i m} frac{e^{-t}}{t-a} , {rm d}t right| leq frac{2pi m}{|a|^2} left( 1+frac{2}{|a|} right) , ?$$
$endgroup$
– Diger
Jan 30 at 0:00
$begingroup$
@Diger I do expect a better bound, since it seems clear that, for fixed $a$, the limit as $mathfrak{Re} (m) to infty$ of the real part of the integral converges (or is of the form $C + O ( frac{1}{m} ) $ for a constant $C$). I would like to upper and lower bound this result as $m$ becomes larger
$endgroup$
– user3141592
Jan 30 at 9:35
$begingroup$
@Diger I do expect a better bound, since it seems clear that, for fixed $a$, the limit as $mathfrak{Re} (m) to infty$ of the real part of the integral converges (or is of the form $C + O ( frac{1}{m} ) $ for a constant $C$). I would like to upper and lower bound this result as $m$ becomes larger
$endgroup$
– user3141592
Jan 30 at 9:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I can't solve or bound the integral for finite $m$, but I can give you a solution for $m to infty$ using Fourier Transforms.
Some notation conventions I'll use
$$mathscr{F}left{f(x)right} = F(s) = int_{-infty}^{infty} {f(x)e^{-2pi i sx} }dx$$
$$ mathrm{sinc}(x) = dfrac{sin(pi x)}{pi x}$$
$$ Pi(x) = begin{cases} 1 quad |x|<frac{1}{2} \
0 quad |x|>frac{1}{2}\
end{cases}$$
$$ H(x) = begin{cases} 0 quad x < 0 \
1 quad x > 0\
end{cases}$$
I'll also use this Fourier Transform pair
$$mathscr{F}left{dfrac{1}{x-z_0}right} = -ipi e^{-2pi i s z_0}left[mathrm{sgn}(s)-mathrm{sgn}left(Imleft[z_0right]right)right]$$
Starting from your integral
$$begin{align*} f(m,a) & =int_0^{2 pi i m} frac{e^{-t}}{t-a} dt\
\
&= 2pi i m int_0^1 {dfrac{e^{-2pi i mx}}{2pi i mx - a}}dx\
\
&= int_0^1{dfrac{e^{-2pi i mx}}{x - dfrac{a}{2pi i m}}}dx\
\
&= int_{-infty}^{infty}dfrac{1}{x - dfrac{a}{2pi i m}}Pileft(x-frac{1}{2}right)e^{-2pi i mx}dx\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)right}\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}right} * mathscr{F}left{Pileft(x-frac{1}{2}right)right}\
\
&= -ipi e^{-2pi i m frac{a}{2pi i m}}left[mathrm{sgn}(m)-mathrm{sgn}left(Imleft[dfrac{a}{2pi i m}right]right)right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1-mathrm{sgn}left(Imleft[dfrac{-ia}{2pi}right]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2H(m)-1right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]int_{-infty}^infty left[2H(m-tau)-1right] e^{-ipi tau}mathrm{sinc}(tau)space dtau \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^infty H(m-tau) e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtauright] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-2pi is tau}mathrm{sinc}(tau)space dtaubiggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - Pi(s)biggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{Pileft({frac{1}{2}}^-right)+ Pileft({frac{1}{2}}^+right)}{2}right] \
\
&= -ipi e^{-Re(a)}e^{-iIm(a)}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
end{align*}$$
If $Re(a) < 0$, then $f(m,a) = 0$.
Also $lim_{Re(a) to infty} f(m,a) = 0$.
For $Re(a) ge 0$, as $m to infty$
$$begin{align*}lim_{m to infty} f(m,a) &= lim_{m to infty} left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2Pi(s)|_{s=frac{1}{2}} - dfrac{1}{2}right] \
\
&= dfrac{left[1+mathrm{sgn}left(Releft[aright]right)right]}{2}pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right] \
\
end{align*}$$
which should make obvious the value to which $Re[f(m,a)]$ is converging.
Unfortunately, the integral
$$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$
is problematic for finite $m$. The real part converges rapidly to $frac{1}{2}$. However, I suspect its imaginary part diverges because $Imleft[frac{i log(-infty)}{2pi}right] to infty$. So I don't know how to compute it for finite $m$.
answered Feb 3 at 21:35
Andy WallsAndy Walls
1,794139
$endgroup$
$begingroup$
This seems helpful. I will take a deeper look at it during these days, thank you!
$endgroup$
– user3141592
Feb 4 at 15:06
$begingroup$
You're welcome. If it helps, Wolfram Alpha comes up with a "closed" form for the anti-derivative for that problematic integrand: wolframalpha.com/input/…
$endgroup$
– Andy Walls
Feb 4 at 19:47
$begingroup$
Well, this result seems not to be correct, since $$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$ does diverge
$endgroup$
– user3141592
Feb 4 at 20:56
$begingroup$
Well, in fact the problem comes from the step where you apply the convolution theorem, since the imaginary part of the convolucion of the Fourier Transform of both functions does not converge (the real part does).
$endgroup$
– user3141592
Feb 4 at 21:24
$begingroup$
On the other hand, this approach can be useful, since, as $$dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)$$ is absolutely integrable, the existence of its Fourier Transform is guaranteed
$endgroup$
– user3141592
Feb 4 at 21:34
|
show 5 more comments
$begingroup$
Assume that $a$ is fixed and has a positive real part. Take the incomplete gamma function $Gamma(0, z)$ to have the branch cut along the negative real axis and to be continuous from above on the branch cut. Then
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} int_{-a}^{-a + 2 pi i m} frac {e^{-t}} t dt = \
e^{-a} (Gamma(0, -a) - Gamma(0, -a + 2 pi i m) -
2 pi i ,[operatorname{Im} a > 0]),$$
since the antiderivative $-Gamma(0, t)$ has a $2 pi i$ jump if the segment $[-a, -a + 2 pi i m]$ crosses the branch cut.
Then, from known asymptotics of $Gamma(0, z)$,
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} (Gamma(0, -a) - 2 pi i ,[operatorname{Im} a > 0]) -
frac 1 {2 pi i m} + O {left( frac 1 {m^2} right)}, \
m to infty, ; m in mathbb N.$$
The real part will not contain the $1/(2 pi i m)$ term.
answered Feb 21 at 23:17
MaximMaxim
6,2131221
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092030%2fsolving-or-bounding-the-real-part-of-the-integral-int-02-pi-i-m-frace%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I can't solve or bound the integral for finite $m$, but I can give you a solution for $m to infty$ using Fourier Transforms.
Some notation conventions I'll use
$$mathscr{F}left{f(x)right} = F(s) = int_{-infty}^{infty} {f(x)e^{-2pi i sx} }dx$$
$$ mathrm{sinc}(x) = dfrac{sin(pi x)}{pi x}$$
$$ Pi(x) = begin{cases} 1 quad |x|<frac{1}{2} \
0 quad |x|>frac{1}{2}\
end{cases}$$
$$ H(x) = begin{cases} 0 quad x < 0 \
1 quad x > 0\
end{cases}$$
I'll also use this Fourier Transform pair
$$mathscr{F}left{dfrac{1}{x-z_0}right} = -ipi e^{-2pi i s z_0}left[mathrm{sgn}(s)-mathrm{sgn}left(Imleft[z_0right]right)right]$$
Starting from your integral
$$begin{align*} f(m,a) & =int_0^{2 pi i m} frac{e^{-t}}{t-a} dt\
\
&= 2pi i m int_0^1 {dfrac{e^{-2pi i mx}}{2pi i mx - a}}dx\
\
&= int_0^1{dfrac{e^{-2pi i mx}}{x - dfrac{a}{2pi i m}}}dx\
\
&= int_{-infty}^{infty}dfrac{1}{x - dfrac{a}{2pi i m}}Pileft(x-frac{1}{2}right)e^{-2pi i mx}dx\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)right}\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}right} * mathscr{F}left{Pileft(x-frac{1}{2}right)right}\
\
&= -ipi e^{-2pi i m frac{a}{2pi i m}}left[mathrm{sgn}(m)-mathrm{sgn}left(Imleft[dfrac{a}{2pi i m}right]right)right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1-mathrm{sgn}left(Imleft[dfrac{-ia}{2pi}right]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2H(m)-1right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]int_{-infty}^infty left[2H(m-tau)-1right] e^{-ipi tau}mathrm{sinc}(tau)space dtau \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^infty H(m-tau) e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtauright] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-2pi is tau}mathrm{sinc}(tau)space dtaubiggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - Pi(s)biggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{Pileft({frac{1}{2}}^-right)+ Pileft({frac{1}{2}}^+right)}{2}right] \
\
&= -ipi e^{-Re(a)}e^{-iIm(a)}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
end{align*}$$
If $Re(a) < 0$, then $f(m,a) = 0$.
Also $lim_{Re(a) to infty} f(m,a) = 0$.
For $Re(a) ge 0$, as $m to infty$
$$begin{align*}lim_{m to infty} f(m,a) &= lim_{m to infty} left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2Pi(s)|_{s=frac{1}{2}} - dfrac{1}{2}right] \
\
&= dfrac{left[1+mathrm{sgn}left(Releft[aright]right)right]}{2}pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right] \
\
end{align*}$$
which should make obvious the value to which $Re[f(m,a)]$ is converging.
Unfortunately, the integral
$$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$
is problematic for finite $m$. The real part converges rapidly to $frac{1}{2}$. However, I suspect its imaginary part diverges because $Imleft[frac{i log(-infty)}{2pi}right] to infty$. So I don't know how to compute it for finite $m$.
answered Feb 3 at 21:35
Andy WallsAndy Walls
1,794139
$endgroup$
$begingroup$
This seems helpful. I will take a deeper look at it during these days, thank you!
$endgroup$
– user3141592
Feb 4 at 15:06
$begingroup$
You're welcome. If it helps, Wolfram Alpha comes up with a "closed" form for the anti-derivative for that problematic integrand: wolframalpha.com/input/…
$endgroup$
– Andy Walls
Feb 4 at 19:47
$begingroup$
Well, this result seems not to be correct, since $$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$ does diverge
$endgroup$
– user3141592
Feb 4 at 20:56
$begingroup$
Well, in fact the problem comes from the step where you apply the convolution theorem, since the imaginary part of the convolucion of the Fourier Transform of both functions does not converge (the real part does).
$endgroup$
– user3141592
Feb 4 at 21:24
$begingroup$
On the other hand, this approach can be useful, since, as $$dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)$$ is absolutely integrable, the existence of its Fourier Transform is guaranteed
$endgroup$
– user3141592
Feb 4 at 21:34
|
show 5 more comments
$begingroup$
I can't solve or bound the integral for finite $m$, but I can give you a solution for $m to infty$ using Fourier Transforms.
Some notation conventions I'll use
$$mathscr{F}left{f(x)right} = F(s) = int_{-infty}^{infty} {f(x)e^{-2pi i sx} }dx$$
$$ mathrm{sinc}(x) = dfrac{sin(pi x)}{pi x}$$
$$ Pi(x) = begin{cases} 1 quad |x|<frac{1}{2} \
0 quad |x|>frac{1}{2}\
end{cases}$$
$$ H(x) = begin{cases} 0 quad x < 0 \
1 quad x > 0\
end{cases}$$
I'll also use this Fourier Transform pair
$$mathscr{F}left{dfrac{1}{x-z_0}right} = -ipi e^{-2pi i s z_0}left[mathrm{sgn}(s)-mathrm{sgn}left(Imleft[z_0right]right)right]$$
Starting from your integral
$$begin{align*} f(m,a) & =int_0^{2 pi i m} frac{e^{-t}}{t-a} dt\
\
&= 2pi i m int_0^1 {dfrac{e^{-2pi i mx}}{2pi i mx - a}}dx\
\
&= int_0^1{dfrac{e^{-2pi i mx}}{x - dfrac{a}{2pi i m}}}dx\
\
&= int_{-infty}^{infty}dfrac{1}{x - dfrac{a}{2pi i m}}Pileft(x-frac{1}{2}right)e^{-2pi i mx}dx\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)right}\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}right} * mathscr{F}left{Pileft(x-frac{1}{2}right)right}\
\
&= -ipi e^{-2pi i m frac{a}{2pi i m}}left[mathrm{sgn}(m)-mathrm{sgn}left(Imleft[dfrac{a}{2pi i m}right]right)right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1-mathrm{sgn}left(Imleft[dfrac{-ia}{2pi}right]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2H(m)-1right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]int_{-infty}^infty left[2H(m-tau)-1right] e^{-ipi tau}mathrm{sinc}(tau)space dtau \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^infty H(m-tau) e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtauright] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-2pi is tau}mathrm{sinc}(tau)space dtaubiggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - Pi(s)biggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{Pileft({frac{1}{2}}^-right)+ Pileft({frac{1}{2}}^+right)}{2}right] \
\
&= -ipi e^{-Re(a)}e^{-iIm(a)}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
end{align*}$$
If $Re(a) < 0$, then $f(m,a) = 0$.
Also $lim_{Re(a) to infty} f(m,a) = 0$.
For $Re(a) ge 0$, as $m to infty$
$$begin{align*}lim_{m to infty} f(m,a) &= lim_{m to infty} left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2Pi(s)|_{s=frac{1}{2}} - dfrac{1}{2}right] \
\
&= dfrac{left[1+mathrm{sgn}left(Releft[aright]right)right]}{2}pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right] \
\
end{align*}$$
which should make obvious the value to which $Re[f(m,a)]$ is converging.
Unfortunately, the integral
$$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$
is problematic for finite $m$. The real part converges rapidly to $frac{1}{2}$. However, I suspect its imaginary part diverges because $Imleft[frac{i log(-infty)}{2pi}right] to infty$. So I don't know how to compute it for finite $m$.
answered Feb 3 at 21:35
Andy WallsAndy Walls
1,794139
$endgroup$
$begingroup$
This seems helpful. I will take a deeper look at it during these days, thank you!
$endgroup$
– user3141592
Feb 4 at 15:06
$begingroup$
You're welcome. If it helps, Wolfram Alpha comes up with a "closed" form for the anti-derivative for that problematic integrand: wolframalpha.com/input/…
$endgroup$
– Andy Walls
Feb 4 at 19:47
$begingroup$
Well, this result seems not to be correct, since $$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$ does diverge
$endgroup$
– user3141592
Feb 4 at 20:56
$begingroup$
Well, in fact the problem comes from the step where you apply the convolution theorem, since the imaginary part of the convolucion of the Fourier Transform of both functions does not converge (the real part does).
$endgroup$
– user3141592
Feb 4 at 21:24
$begingroup$
On the other hand, this approach can be useful, since, as $$dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)$$ is absolutely integrable, the existence of its Fourier Transform is guaranteed
$endgroup$
– user3141592
Feb 4 at 21:34
|
show 5 more comments
$begingroup$
I can't solve or bound the integral for finite $m$, but I can give you a solution for $m to infty$ using Fourier Transforms.
Some notation conventions I'll use
$$mathscr{F}left{f(x)right} = F(s) = int_{-infty}^{infty} {f(x)e^{-2pi i sx} }dx$$
$$ mathrm{sinc}(x) = dfrac{sin(pi x)}{pi x}$$
$$ Pi(x) = begin{cases} 1 quad |x|<frac{1}{2} \
0 quad |x|>frac{1}{2}\
end{cases}$$
$$ H(x) = begin{cases} 0 quad x < 0 \
1 quad x > 0\
end{cases}$$
I'll also use this Fourier Transform pair
$$mathscr{F}left{dfrac{1}{x-z_0}right} = -ipi e^{-2pi i s z_0}left[mathrm{sgn}(s)-mathrm{sgn}left(Imleft[z_0right]right)right]$$
Starting from your integral
$$begin{align*} f(m,a) & =int_0^{2 pi i m} frac{e^{-t}}{t-a} dt\
\
&= 2pi i m int_0^1 {dfrac{e^{-2pi i mx}}{2pi i mx - a}}dx\
\
&= int_0^1{dfrac{e^{-2pi i mx}}{x - dfrac{a}{2pi i m}}}dx\
\
&= int_{-infty}^{infty}dfrac{1}{x - dfrac{a}{2pi i m}}Pileft(x-frac{1}{2}right)e^{-2pi i mx}dx\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)right}\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}right} * mathscr{F}left{Pileft(x-frac{1}{2}right)right}\
\
&= -ipi e^{-2pi i m frac{a}{2pi i m}}left[mathrm{sgn}(m)-mathrm{sgn}left(Imleft[dfrac{a}{2pi i m}right]right)right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1-mathrm{sgn}left(Imleft[dfrac{-ia}{2pi}right]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2H(m)-1right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]int_{-infty}^infty left[2H(m-tau)-1right] e^{-ipi tau}mathrm{sinc}(tau)space dtau \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^infty H(m-tau) e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtauright] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-2pi is tau}mathrm{sinc}(tau)space dtaubiggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - Pi(s)biggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{Pileft({frac{1}{2}}^-right)+ Pileft({frac{1}{2}}^+right)}{2}right] \
\
&= -ipi e^{-Re(a)}e^{-iIm(a)}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
end{align*}$$
If $Re(a) < 0$, then $f(m,a) = 0$.
Also $lim_{Re(a) to infty} f(m,a) = 0$.
For $Re(a) ge 0$, as $m to infty$
$$begin{align*}lim_{m to infty} f(m,a) &= lim_{m to infty} left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2Pi(s)|_{s=frac{1}{2}} - dfrac{1}{2}right] \
\
&= dfrac{left[1+mathrm{sgn}left(Releft[aright]right)right]}{2}pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right] \
\
end{align*}$$
which should make obvious the value to which $Re[f(m,a)]$ is converging.
Unfortunately, the integral
$$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$
is problematic for finite $m$. The real part converges rapidly to $frac{1}{2}$. However, I suspect its imaginary part diverges because $Imleft[frac{i log(-infty)}{2pi}right] to infty$. So I don't know how to compute it for finite $m$.
answered Feb 3 at 21:35
Andy WallsAndy Walls
1,794139
$endgroup$
I can't solve or bound the integral for finite $m$, but I can give you a solution for $m to infty$ using Fourier Transforms.
Some notation conventions I'll use
$$mathscr{F}left{f(x)right} = F(s) = int_{-infty}^{infty} {f(x)e^{-2pi i sx} }dx$$
$$ mathrm{sinc}(x) = dfrac{sin(pi x)}{pi x}$$
$$ Pi(x) = begin{cases} 1 quad |x|<frac{1}{2} \
0 quad |x|>frac{1}{2}\
end{cases}$$
$$ H(x) = begin{cases} 0 quad x < 0 \
1 quad x > 0\
end{cases}$$
I'll also use this Fourier Transform pair
$$mathscr{F}left{dfrac{1}{x-z_0}right} = -ipi e^{-2pi i s z_0}left[mathrm{sgn}(s)-mathrm{sgn}left(Imleft[z_0right]right)right]$$
Starting from your integral
$$begin{align*} f(m,a) & =int_0^{2 pi i m} frac{e^{-t}}{t-a} dt\
\
&= 2pi i m int_0^1 {dfrac{e^{-2pi i mx}}{2pi i mx - a}}dx\
\
&= int_0^1{dfrac{e^{-2pi i mx}}{x - dfrac{a}{2pi i m}}}dx\
\
&= int_{-infty}^{infty}dfrac{1}{x - dfrac{a}{2pi i m}}Pileft(x-frac{1}{2}right)e^{-2pi i mx}dx\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)right}\
\
&= mathscr{F}left{dfrac{1}{x - dfrac{a}{2pi i m}}right} * mathscr{F}left{Pileft(x-frac{1}{2}right)right}\
\
&= -ipi e^{-2pi i m frac{a}{2pi i m}}left[mathrm{sgn}(m)-mathrm{sgn}left(Imleft[dfrac{a}{2pi i m}right]right)right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1-mathrm{sgn}left(Imleft[dfrac{-ia}{2pi}right]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]mathrm{sgn}(m) * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2H(m)-1right] * e^{-ipi m}mathrm{sinc}(m)\
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]int_{-infty}^infty left[2H(m-tau)-1right] e^{-ipi tau}mathrm{sinc}(tau)space dtau \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^infty H(m-tau) e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtauright] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - int_{-infty}^infty e^{-2pi is tau}mathrm{sinc}(tau)space dtaubiggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - Pi(s)biggr|_{s=frac{1}{2}}right] \
\
&= -ipi e^{-a}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{Pileft({frac{1}{2}}^-right)+ Pileft({frac{1}{2}}^+right)}{2}right] \
\
&= -ipi e^{-Re(a)}e^{-iIm(a)}left[1+mathrm{sgn}left(Releft[aright]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
end{align*}$$
If $Re(a) < 0$, then $f(m,a) = 0$.
Also $lim_{Re(a) to infty} f(m,a) = 0$.
For $Re(a) ge 0$, as $m to infty$
$$begin{align*}lim_{m to infty} f(m,a) &= lim_{m to infty} left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2int_{-infty}^infty e^{-ipi tau}mathrm{sinc}(tau)space dtau - dfrac{1}{2}right] \
\
&= left[1+mathrm{sgn}left(Releft[aright]right)right]pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right]left[2Pi(s)|_{s=frac{1}{2}} - dfrac{1}{2}right] \
\
&= dfrac{left[1+mathrm{sgn}left(Releft[aright]right)right]}{2}pi e^{-Re(a)}left[-sinleft(Im[a]right)-icosleft(Im[a]right)right] \
\
end{align*}$$
which should make obvious the value to which $Re[f(m,a)]$ is converging.
Unfortunately, the integral
$$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$
is problematic for finite $m$. The real part converges rapidly to $frac{1}{2}$. However, I suspect its imaginary part diverges because $Imleft[frac{i log(-infty)}{2pi}right] to infty$. So I don't know how to compute it for finite $m$.
answered Feb 3 at 21:35
Andy WallsAndy Walls
1,794139
edited Feb 4 at 19:49
answered Feb 3 at 21:35
Andy WallsAndy Walls
1,794139
answered Feb 3 at 21:35
Andy WallsAndy Walls
1,794139
answered Feb 3 at 21:35
Andy WallsAndy Walls
1,794139
1,794139
$begingroup$
This seems helpful. I will take a deeper look at it during these days, thank you!
$endgroup$
– user3141592
Feb 4 at 15:06
$begingroup$
You're welcome. If it helps, Wolfram Alpha comes up with a "closed" form for the anti-derivative for that problematic integrand: wolframalpha.com/input/…
$endgroup$
– Andy Walls
Feb 4 at 19:47
$begingroup$
Well, this result seems not to be correct, since $$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$ does diverge
$endgroup$
– user3141592
Feb 4 at 20:56
$begingroup$
Well, in fact the problem comes from the step where you apply the convolution theorem, since the imaginary part of the convolucion of the Fourier Transform of both functions does not converge (the real part does).
$endgroup$
– user3141592
Feb 4 at 21:24
$begingroup$
On the other hand, this approach can be useful, since, as $$dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)$$ is absolutely integrable, the existence of its Fourier Transform is guaranteed
$endgroup$
– user3141592
Feb 4 at 21:34
|
show 5 more comments
$begingroup$
This seems helpful. I will take a deeper look at it during these days, thank you!
$endgroup$
– user3141592
Feb 4 at 15:06
$begingroup$
You're welcome. If it helps, Wolfram Alpha comes up with a "closed" form for the anti-derivative for that problematic integrand: wolframalpha.com/input/…
$endgroup$
– Andy Walls
Feb 4 at 19:47
$begingroup$
Well, this result seems not to be correct, since $$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$ does diverge
$endgroup$
– user3141592
Feb 4 at 20:56
$begingroup$
Well, in fact the problem comes from the step where you apply the convolution theorem, since the imaginary part of the convolucion of the Fourier Transform of both functions does not converge (the real part does).
$endgroup$
– user3141592
Feb 4 at 21:24
$begingroup$
On the other hand, this approach can be useful, since, as $$dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)$$ is absolutely integrable, the existence of its Fourier Transform is guaranteed
$endgroup$
– user3141592
Feb 4 at 21:34
$begingroup$
This seems helpful. I will take a deeper look at it during these days, thank you!
$endgroup$
– user3141592
Feb 4 at 15:06
$begingroup$
This seems helpful. I will take a deeper look at it during these days, thank you!
$endgroup$
– user3141592
Feb 4 at 15:06
$begingroup$
You're welcome. If it helps, Wolfram Alpha comes up with a "closed" form for the anti-derivative for that problematic integrand: wolframalpha.com/input/…
$endgroup$
– Andy Walls
Feb 4 at 19:47
$begingroup$
You're welcome. If it helps, Wolfram Alpha comes up with a "closed" form for the anti-derivative for that problematic integrand: wolframalpha.com/input/…
$endgroup$
– Andy Walls
Feb 4 at 19:47
$begingroup$
Well, this result seems not to be correct, since $$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$ does diverge
$endgroup$
– user3141592
Feb 4 at 20:56
$begingroup$
Well, this result seems not to be correct, since $$int_{-infty}^m e^{-ipi tau}mathrm{sinc}(tau)space dtau$$ does diverge
$endgroup$
– user3141592
Feb 4 at 20:56
$begingroup$
Well, in fact the problem comes from the step where you apply the convolution theorem, since the imaginary part of the convolucion of the Fourier Transform of both functions does not converge (the real part does).
$endgroup$
– user3141592
Feb 4 at 21:24
$begingroup$
Well, in fact the problem comes from the step where you apply the convolution theorem, since the imaginary part of the convolucion of the Fourier Transform of both functions does not converge (the real part does).
$endgroup$
– user3141592
Feb 4 at 21:24
$begingroup$
On the other hand, this approach can be useful, since, as $$dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)$$ is absolutely integrable, the existence of its Fourier Transform is guaranteed
$endgroup$
– user3141592
Feb 4 at 21:34
$begingroup$
On the other hand, this approach can be useful, since, as $$dfrac{1}{x - dfrac{a}{2pi i m}}cdotPileft(x-frac{1}{2}right)$$ is absolutely integrable, the existence of its Fourier Transform is guaranteed
$endgroup$
– user3141592
Feb 4 at 21:34
|
show 5 more comments
$begingroup$
Assume that $a$ is fixed and has a positive real part. Take the incomplete gamma function $Gamma(0, z)$ to have the branch cut along the negative real axis and to be continuous from above on the branch cut. Then
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} int_{-a}^{-a + 2 pi i m} frac {e^{-t}} t dt = \
e^{-a} (Gamma(0, -a) - Gamma(0, -a + 2 pi i m) -
2 pi i ,[operatorname{Im} a > 0]),$$
since the antiderivative $-Gamma(0, t)$ has a $2 pi i$ jump if the segment $[-a, -a + 2 pi i m]$ crosses the branch cut.
Then, from known asymptotics of $Gamma(0, z)$,
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} (Gamma(0, -a) - 2 pi i ,[operatorname{Im} a > 0]) -
frac 1 {2 pi i m} + O {left( frac 1 {m^2} right)}, \
m to infty, ; m in mathbb N.$$
The real part will not contain the $1/(2 pi i m)$ term.
answered Feb 21 at 23:17
MaximMaxim
6,2131221
$endgroup$
add a comment |
$begingroup$
Assume that $a$ is fixed and has a positive real part. Take the incomplete gamma function $Gamma(0, z)$ to have the branch cut along the negative real axis and to be continuous from above on the branch cut. Then
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} int_{-a}^{-a + 2 pi i m} frac {e^{-t}} t dt = \
e^{-a} (Gamma(0, -a) - Gamma(0, -a + 2 pi i m) -
2 pi i ,[operatorname{Im} a > 0]),$$
since the antiderivative $-Gamma(0, t)$ has a $2 pi i$ jump if the segment $[-a, -a + 2 pi i m]$ crosses the branch cut.
Then, from known asymptotics of $Gamma(0, z)$,
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} (Gamma(0, -a) - 2 pi i ,[operatorname{Im} a > 0]) -
frac 1 {2 pi i m} + O {left( frac 1 {m^2} right)}, \
m to infty, ; m in mathbb N.$$
The real part will not contain the $1/(2 pi i m)$ term.
answered Feb 21 at 23:17
MaximMaxim
6,2131221
$endgroup$
add a comment |
$begingroup$
Assume that $a$ is fixed and has a positive real part. Take the incomplete gamma function $Gamma(0, z)$ to have the branch cut along the negative real axis and to be continuous from above on the branch cut. Then
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} int_{-a}^{-a + 2 pi i m} frac {e^{-t}} t dt = \
e^{-a} (Gamma(0, -a) - Gamma(0, -a + 2 pi i m) -
2 pi i ,[operatorname{Im} a > 0]),$$
since the antiderivative $-Gamma(0, t)$ has a $2 pi i$ jump if the segment $[-a, -a + 2 pi i m]$ crosses the branch cut.
Then, from known asymptotics of $Gamma(0, z)$,
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} (Gamma(0, -a) - 2 pi i ,[operatorname{Im} a > 0]) -
frac 1 {2 pi i m} + O {left( frac 1 {m^2} right)}, \
m to infty, ; m in mathbb N.$$
The real part will not contain the $1/(2 pi i m)$ term.
answered Feb 21 at 23:17
MaximMaxim
6,2131221
$endgroup$
Assume that $a$ is fixed and has a positive real part. Take the incomplete gamma function $Gamma(0, z)$ to have the branch cut along the negative real axis and to be continuous from above on the branch cut. Then
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} int_{-a}^{-a + 2 pi i m} frac {e^{-t}} t dt = \
e^{-a} (Gamma(0, -a) - Gamma(0, -a + 2 pi i m) -
2 pi i ,[operatorname{Im} a > 0]),$$
since the antiderivative $-Gamma(0, t)$ has a $2 pi i$ jump if the segment $[-a, -a + 2 pi i m]$ crosses the branch cut.
Then, from known asymptotics of $Gamma(0, z)$,
$$int_0^{2 pi i m} frac {e^{-t}} {t - a} dt =
e^{-a} (Gamma(0, -a) - 2 pi i ,[operatorname{Im} a > 0]) -
frac 1 {2 pi i m} + O {left( frac 1 {m^2} right)}, \
m to infty, ; m in mathbb N.$$
The real part will not contain the $1/(2 pi i m)$ term.
answered Feb 21 at 23:17
MaximMaxim
6,2131221
answered Feb 21 at 23:17
MaximMaxim
6,2131221
answered Feb 21 at 23:17
MaximMaxim
6,2131221
answered Feb 21 at 23:17
MaximMaxim
6,2131221
6,2131221
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092030%2fsolving-or-bounding-the-real-part-of-the-integral-int-02-pi-i-m-frace%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
What do you mean by bounding? Something like $$left| int_0^{2 pi i m} frac{e^{-t}}{t-a} , {rm d}t right| leq frac{2pi m}{|a|^2} left( 1+frac{2}{|a|} right) , ?$$
$endgroup$
– Diger
Jan 30 at 0:00
$begingroup$
@Diger I do expect a better bound, since it seems clear that, for fixed $a$, the limit as $mathfrak{Re} (m) to infty$ of the real part of the integral converges (or is of the form $C + O ( frac{1}{m} ) $ for a constant $C$). I would like to upper and lower bound this result as $m$ becomes larger
$endgroup$
– user3141592
Jan 30 at 9:35