Mixing
Cauchy Principal Value of $int_{-infty}^{infty} dfrac{sin x}{x(x^2+1)}mathrm dx$
$begingroup$
How to evaluate this integral
$$
int_{-infty}^inftyfrac{sin x}{x(x^2 + 1)}dx
$$
I am having a problem to solve this because of two poles when I solve it by integration first from $-R$ to $R$ and then along a semicircle in the upper half plane.
complex-analysis
edited Jan 25 at 17:51
Paras Khosla
2,258222
asked Jan 25 at 14:23
sweety tarikasweety tarika
255
$endgroup$
add a comment |
$begingroup$
How to evaluate this integral
$$
int_{-infty}^inftyfrac{sin x}{x(x^2 + 1)}dx
$$
I am having a problem to solve this because of two poles when I solve it by integration first from $-R$ to $R$ and then along a semicircle in the upper half plane.
complex-analysis
edited Jan 25 at 17:51
Paras Khosla
2,258222
asked Jan 25 at 14:23
sweety tarikasweety tarika
255
$endgroup$
1
$begingroup$
Use Residue theorem. Note that only $0$ and $i$ lie in the upper half plane.
$endgroup$
– Paras Khosla
Jan 25 at 14:35
$begingroup$
You need to show your work, so that you can get direction towards the approach that'll get you to the solution.
$endgroup$
– Paras Khosla
Jan 25 at 14:50
$begingroup$
And when i try to evalute ∫e∧iz/ z(z²⁺1) problem is that z=0 pole comes on boundry of our closed contour
$endgroup$
– sweety tarika
Jan 25 at 15:35
$begingroup$
$z=0$ should not be a problem because in fact $text{Res} bigl(frac{sin z}{z(z^2+1)}, 0bigr)=0$
$endgroup$
– Paras Khosla
Jan 25 at 16:06
$begingroup$
see z=0 is not inside the contour...its on boundry
$endgroup$
– sweety tarika
Jan 25 at 16:35
add a comment |
$begingroup$
How to evaluate this integral
$$
int_{-infty}^inftyfrac{sin x}{x(x^2 + 1)}dx
$$
I am having a problem to solve this because of two poles when I solve it by integration first from $-R$ to $R$ and then along a semicircle in the upper half plane.
complex-analysis
edited Jan 25 at 17:51
Paras Khosla
2,258222
asked Jan 25 at 14:23
sweety tarikasweety tarika
255
$endgroup$
How to evaluate this integral
$$
int_{-infty}^inftyfrac{sin x}{x(x^2 + 1)}dx
$$
I am having a problem to solve this because of two poles when I solve it by integration first from $-R$ to $R$ and then along a semicircle in the upper half plane.
complex-analysis
complex-analysis
edited Jan 25 at 17:51
Paras Khosla
2,258222
asked Jan 25 at 14:23
sweety tarikasweety tarika
255
edited Jan 25 at 17:51
Paras Khosla
2,258222
asked Jan 25 at 14:23
sweety tarikasweety tarika
255
edited Jan 25 at 17:51
Paras Khosla
2,258222
edited Jan 25 at 17:51
Paras Khosla
2,258222
edited Jan 25 at 17:51
Paras Khosla
2,258222
2,258222
asked Jan 25 at 14:23
sweety tarikasweety tarika
255
asked Jan 25 at 14:23
sweety tarikasweety tarika
255
asked Jan 25 at 14:23
sweety tarikasweety tarika
255
255
1
$begingroup$
Use Residue theorem. Note that only $0$ and $i$ lie in the upper half plane.
$endgroup$
– Paras Khosla
Jan 25 at 14:35
$begingroup$
You need to show your work, so that you can get direction towards the approach that'll get you to the solution.
$endgroup$
– Paras Khosla
Jan 25 at 14:50
$begingroup$
And when i try to evalute ∫e∧iz/ z(z²⁺1) problem is that z=0 pole comes on boundry of our closed contour
$endgroup$
– sweety tarika
Jan 25 at 15:35
$begingroup$
$z=0$ should not be a problem because in fact $text{Res} bigl(frac{sin z}{z(z^2+1)}, 0bigr)=0$
$endgroup$
– Paras Khosla
Jan 25 at 16:06
$begingroup$
see z=0 is not inside the contour...its on boundry
$endgroup$
– sweety tarika
Jan 25 at 16:35
add a comment |
1
$begingroup$
Use Residue theorem. Note that only $0$ and $i$ lie in the upper half plane.
$endgroup$
– Paras Khosla
Jan 25 at 14:35
$begingroup$
You need to show your work, so that you can get direction towards the approach that'll get you to the solution.
$endgroup$
– Paras Khosla
Jan 25 at 14:50
$begingroup$
And when i try to evalute ∫e∧iz/ z(z²⁺1) problem is that z=0 pole comes on boundry of our closed contour
$endgroup$
– sweety tarika
Jan 25 at 15:35
$begingroup$
$z=0$ should not be a problem because in fact $text{Res} bigl(frac{sin z}{z(z^2+1)}, 0bigr)=0$
$endgroup$
– Paras Khosla
Jan 25 at 16:06
$begingroup$
see z=0 is not inside the contour...its on boundry
$endgroup$
– sweety tarika
Jan 25 at 16:35
1
1
$begingroup$
Use Residue theorem. Note that only $0$ and $i$ lie in the upper half plane.
$endgroup$
– Paras Khosla
Jan 25 at 14:35
$begingroup$
Use Residue theorem. Note that only $0$ and $i$ lie in the upper half plane.
$endgroup$
– Paras Khosla
Jan 25 at 14:35
$begingroup$
You need to show your work, so that you can get direction towards the approach that'll get you to the solution.
$endgroup$
– Paras Khosla
Jan 25 at 14:50
$begingroup$
You need to show your work, so that you can get direction towards the approach that'll get you to the solution.
$endgroup$
– Paras Khosla
Jan 25 at 14:50
$begingroup$
And when i try to evalute ∫e∧iz/ z(z²⁺1) problem is that z=0 pole comes on boundry of our closed contour
$endgroup$
– sweety tarika
Jan 25 at 15:35
$begingroup$
And when i try to evalute ∫e∧iz/ z(z²⁺1) problem is that z=0 pole comes on boundry of our closed contour
$endgroup$
– sweety tarika
Jan 25 at 15:35
$begingroup$
$z=0$ should not be a problem because in fact $text{Res} bigl(frac{sin z}{z(z^2+1)}, 0bigr)=0$
$endgroup$
– Paras Khosla
Jan 25 at 16:06
$begingroup$
$z=0$ should not be a problem because in fact $text{Res} bigl(frac{sin z}{z(z^2+1)}, 0bigr)=0$
$endgroup$
– Paras Khosla
Jan 25 at 16:06
$begingroup$
see z=0 is not inside the contour...its on boundry
$endgroup$
– sweety tarika
Jan 25 at 16:35
$begingroup$
see z=0 is not inside the contour...its on boundry
$endgroup$
– sweety tarika
Jan 25 at 16:35
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
According to your choice of the contour: $C$ is the upper half-plane and $Gamma$ is the semicircular arc of radius $R$ (say).
$$int_C frac{sin z}{z(z^2+1)}mathrm dz=text{P.V.} int_{-infty}^{+infty} frac{sin x}{x(x^2+1)}mathrm dx+lim_{R to infty}int_{Gamma}frac{sin z}{z(z^2+1)}mathrm dz$$
Now, use the fact that if $f(z)=frac{g(z)}{h(z)}$ where $f$ and $g$ are analytic near $z_0$ and $h$ has a simple zero at $z_0$, then $text{Res}(f(z), z_0) = frac{g(z_0)}{h'(z_0)}$. Note that $g$ is $sin z$ and $h$ is $z(z^2+1)$. For the evaluation of the integral over the arc $Gamma$, use the ML-inequality and you should be good to go.
answered Jan 25 at 15:13
Paras KhoslaParas Khosla
2,258222
$endgroup$
$begingroup$
First of all thanks, but sir we cant use ML inequalty because its unbounded
$endgroup$
– sweety tarika
Jan 25 at 15:30
$begingroup$
No the function is not unbounded. ML Inequality can be used. In fact, $text{P.V.} int_{-infty}^{infty} frac{sin z}{z(z^2+1)}mathrm dz=e^{-1}(e-1)pi$.
$endgroup$
– Paras Khosla
Jan 25 at 16:25
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
i am new user, i dont know how to use this app
$endgroup$
– sweety tarika
Jan 25 at 16:31
|
show 5 more comments
$begingroup$
Let us avoid contour-hunting. The integrand function is even, hence it is enough to compute
$$ F(a)=int_{0}^{+infty}frac{sin(ax)}{x(x^2+1)},dx $$
(defined over $text{Re}(a)>0$) at $a=1$. We may notice that
$$ (mathcal{L} F)(s) = int_{0}^{+infty}frac{dx}{(s^2+x^2)(1+x^2)}=frac{pi}{2s(s+1)} $$
hence
$$F(a) = frac{pi}{2}mathcal{L}^{-1}left(frac{1}{s}-frac{1}{s+1}right)(a)=frac{pi}{2}(1-e^{-a}) $$
and
$$ int_{-infty}^{+infty}frac{sin x}{x(x^2+1)},dx = color{blue}{pi(1-e^{-1})}.$$
There is no need to introduce a principal value: the $text{sinc}$ function is continuous and bounded over $mathbb{R}$.
answered Jan 25 at 17:28
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
$begingroup$
Wowww...its good aproach
$endgroup$
– sweety tarika
Jan 25 at 17:50
$begingroup$
Sir,you directly write value of integration...when i try to solve by putting x² =t ...it become lengthy
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
is there any quick method for this integration?
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
@sweetytarika: quicker than the shown one, you mean? You may use Fourier transforms instead of Laplace transforms, it remains a four-liner or so.
$endgroup$
– Jack D'Aurizio
Jan 25 at 18:10
add a comment |
$begingroup$
Here is another approach that like @Jack D'Aurizio approach avoids contour integration altogether.
Let
$$F(a) = int_{-infty}^infty frac{sin (ax)}{x(x^2 + 1)} , dx = 2 int_0^infty frac{sin (ax)}{x(x^2 + 1)} , dx, qquad a > 0.$$
We are required to find $F(1)$.
Using Feynman's trick of differentiating under the integral sign we have
$$F'(a) = 2 int_0^infty frac{cos (ax)}{x^2 + 1} , dx,$$
and
$$F''(a) = - 2 int_0^infty frac{x sin (ax)}{x^2 + 1} , dx.$$
Observe that
$$F''(a) - F(a) = -2 int_0^infty frac{sin (ax)}{x} , dx = -pi.$$
Here the well-known result of
$$int_0^infty frac{sin (ax)}{x} , dx = frac{pi}{2}, qquad a > 0,$$
has been used.
One solving the differential equaltion $F''(a) - F(a) = -pi$ we have
$$F(a) = C_1 e^a + C_2 e^{-a} + pi.$$
The two unknown constants $C_1$ and $C_2$ can be found by noting that $F(0) = 0$ and $F'(0) = pi$. Doing so one finds
$$F(a) = pi(1 - e^{-a}).$$
Finally, setting $a = 1$ one has
$$int_{-infty}^infty frac{sin x}{x(x^2 + 1)} , dx = pi(1 - e^{-1}),$$
as required.
answered Jan 26 at 6:31
omegadotomegadot
6,4072829
$endgroup$
$begingroup$
in begining how you took F(a⁾=2 ,except this step every step is meaningfull
$endgroup$
– sweety tarika
Jan 26 at 11:40
$begingroup$
If you mean the factor of 2 appearing in front of the integral on the first line, that comes from the integrand being an even function between symmetric limits. Is this what you mean?
$endgroup$
– omegadot
Jan 26 at 11:42
$begingroup$
ohkk, got it..thanks
$endgroup$
– sweety tarika
Jan 26 at 11:49
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to your choice of the contour: $C$ is the upper half-plane and $Gamma$ is the semicircular arc of radius $R$ (say).
$$int_C frac{sin z}{z(z^2+1)}mathrm dz=text{P.V.} int_{-infty}^{+infty} frac{sin x}{x(x^2+1)}mathrm dx+lim_{R to infty}int_{Gamma}frac{sin z}{z(z^2+1)}mathrm dz$$
Now, use the fact that if $f(z)=frac{g(z)}{h(z)}$ where $f$ and $g$ are analytic near $z_0$ and $h$ has a simple zero at $z_0$, then $text{Res}(f(z), z_0) = frac{g(z_0)}{h'(z_0)}$. Note that $g$ is $sin z$ and $h$ is $z(z^2+1)$. For the evaluation of the integral over the arc $Gamma$, use the ML-inequality and you should be good to go.
answered Jan 25 at 15:13
Paras KhoslaParas Khosla
2,258222
$endgroup$
$begingroup$
First of all thanks, but sir we cant use ML inequalty because its unbounded
$endgroup$
– sweety tarika
Jan 25 at 15:30
$begingroup$
No the function is not unbounded. ML Inequality can be used. In fact, $text{P.V.} int_{-infty}^{infty} frac{sin z}{z(z^2+1)}mathrm dz=e^{-1}(e-1)pi$.
$endgroup$
– Paras Khosla
Jan 25 at 16:25
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
i am new user, i dont know how to use this app
$endgroup$
– sweety tarika
Jan 25 at 16:31
|
show 5 more comments
$begingroup$
According to your choice of the contour: $C$ is the upper half-plane and $Gamma$ is the semicircular arc of radius $R$ (say).
$$int_C frac{sin z}{z(z^2+1)}mathrm dz=text{P.V.} int_{-infty}^{+infty} frac{sin x}{x(x^2+1)}mathrm dx+lim_{R to infty}int_{Gamma}frac{sin z}{z(z^2+1)}mathrm dz$$
Now, use the fact that if $f(z)=frac{g(z)}{h(z)}$ where $f$ and $g$ are analytic near $z_0$ and $h$ has a simple zero at $z_0$, then $text{Res}(f(z), z_0) = frac{g(z_0)}{h'(z_0)}$. Note that $g$ is $sin z$ and $h$ is $z(z^2+1)$. For the evaluation of the integral over the arc $Gamma$, use the ML-inequality and you should be good to go.
answered Jan 25 at 15:13
Paras KhoslaParas Khosla
2,258222
$endgroup$
$begingroup$
First of all thanks, but sir we cant use ML inequalty because its unbounded
$endgroup$
– sweety tarika
Jan 25 at 15:30
$begingroup$
No the function is not unbounded. ML Inequality can be used. In fact, $text{P.V.} int_{-infty}^{infty} frac{sin z}{z(z^2+1)}mathrm dz=e^{-1}(e-1)pi$.
$endgroup$
– Paras Khosla
Jan 25 at 16:25
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
i am new user, i dont know how to use this app
$endgroup$
– sweety tarika
Jan 25 at 16:31
|
show 5 more comments
$begingroup$
According to your choice of the contour: $C$ is the upper half-plane and $Gamma$ is the semicircular arc of radius $R$ (say).
$$int_C frac{sin z}{z(z^2+1)}mathrm dz=text{P.V.} int_{-infty}^{+infty} frac{sin x}{x(x^2+1)}mathrm dx+lim_{R to infty}int_{Gamma}frac{sin z}{z(z^2+1)}mathrm dz$$
Now, use the fact that if $f(z)=frac{g(z)}{h(z)}$ where $f$ and $g$ are analytic near $z_0$ and $h$ has a simple zero at $z_0$, then $text{Res}(f(z), z_0) = frac{g(z_0)}{h'(z_0)}$. Note that $g$ is $sin z$ and $h$ is $z(z^2+1)$. For the evaluation of the integral over the arc $Gamma$, use the ML-inequality and you should be good to go.
answered Jan 25 at 15:13
Paras KhoslaParas Khosla
2,258222
$endgroup$
According to your choice of the contour: $C$ is the upper half-plane and $Gamma$ is the semicircular arc of radius $R$ (say).
$$int_C frac{sin z}{z(z^2+1)}mathrm dz=text{P.V.} int_{-infty}^{+infty} frac{sin x}{x(x^2+1)}mathrm dx+lim_{R to infty}int_{Gamma}frac{sin z}{z(z^2+1)}mathrm dz$$
Now, use the fact that if $f(z)=frac{g(z)}{h(z)}$ where $f$ and $g$ are analytic near $z_0$ and $h$ has a simple zero at $z_0$, then $text{Res}(f(z), z_0) = frac{g(z_0)}{h'(z_0)}$. Note that $g$ is $sin z$ and $h$ is $z(z^2+1)$. For the evaluation of the integral over the arc $Gamma$, use the ML-inequality and you should be good to go.
answered Jan 25 at 15:13
Paras KhoslaParas Khosla
2,258222
answered Jan 25 at 15:13
Paras KhoslaParas Khosla
2,258222
answered Jan 25 at 15:13
Paras KhoslaParas Khosla
2,258222
answered Jan 25 at 15:13
Paras KhoslaParas Khosla
2,258222
2,258222
$begingroup$
First of all thanks, but sir we cant use ML inequalty because its unbounded
$endgroup$
– sweety tarika
Jan 25 at 15:30
$begingroup$
No the function is not unbounded. ML Inequality can be used. In fact, $text{P.V.} int_{-infty}^{infty} frac{sin z}{z(z^2+1)}mathrm dz=e^{-1}(e-1)pi$.
$endgroup$
– Paras Khosla
Jan 25 at 16:25
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
i am new user, i dont know how to use this app
$endgroup$
– sweety tarika
Jan 25 at 16:31
|
show 5 more comments
$begingroup$
First of all thanks, but sir we cant use ML inequalty because its unbounded
$endgroup$
– sweety tarika
Jan 25 at 15:30
$begingroup$
No the function is not unbounded. ML Inequality can be used. In fact, $text{P.V.} int_{-infty}^{infty} frac{sin z}{z(z^2+1)}mathrm dz=e^{-1}(e-1)pi$.
$endgroup$
– Paras Khosla
Jan 25 at 16:25
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
i am new user, i dont know how to use this app
$endgroup$
– sweety tarika
Jan 25 at 16:31
$begingroup$
First of all thanks, but sir we cant use ML inequalty because its unbounded
$endgroup$
– sweety tarika
Jan 25 at 15:30
$begingroup$
First of all thanks, but sir we cant use ML inequalty because its unbounded
$endgroup$
– sweety tarika
Jan 25 at 15:30
$begingroup$
No the function is not unbounded. ML Inequality can be used. In fact, $text{P.V.} int_{-infty}^{infty} frac{sin z}{z(z^2+1)}mathrm dz=e^{-1}(e-1)pi$.
$endgroup$
– Paras Khosla
Jan 25 at 16:25
$begingroup$
No the function is not unbounded. ML Inequality can be used. In fact, $text{P.V.} int_{-infty}^{infty} frac{sin z}{z(z^2+1)}mathrm dz=e^{-1}(e-1)pi$.
$endgroup$
– Paras Khosla
Jan 25 at 16:25
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
can you show how you solve?
$endgroup$
– sweety tarika
Jan 25 at 16:27
$begingroup$
i am new user, i dont know how to use this app
$endgroup$
– sweety tarika
Jan 25 at 16:31
$begingroup$
i am new user, i dont know how to use this app
$endgroup$
– sweety tarika
Jan 25 at 16:31
|
show 5 more comments
$begingroup$
Let us avoid contour-hunting. The integrand function is even, hence it is enough to compute
$$ F(a)=int_{0}^{+infty}frac{sin(ax)}{x(x^2+1)},dx $$
(defined over $text{Re}(a)>0$) at $a=1$. We may notice that
$$ (mathcal{L} F)(s) = int_{0}^{+infty}frac{dx}{(s^2+x^2)(1+x^2)}=frac{pi}{2s(s+1)} $$
hence
$$F(a) = frac{pi}{2}mathcal{L}^{-1}left(frac{1}{s}-frac{1}{s+1}right)(a)=frac{pi}{2}(1-e^{-a}) $$
and
$$ int_{-infty}^{+infty}frac{sin x}{x(x^2+1)},dx = color{blue}{pi(1-e^{-1})}.$$
There is no need to introduce a principal value: the $text{sinc}$ function is continuous and bounded over $mathbb{R}$.
answered Jan 25 at 17:28
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
$begingroup$
Wowww...its good aproach
$endgroup$
– sweety tarika
Jan 25 at 17:50
$begingroup$
Sir,you directly write value of integration...when i try to solve by putting x² =t ...it become lengthy
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
is there any quick method for this integration?
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
@sweetytarika: quicker than the shown one, you mean? You may use Fourier transforms instead of Laplace transforms, it remains a four-liner or so.
$endgroup$
– Jack D'Aurizio
Jan 25 at 18:10
add a comment |
$begingroup$
Let us avoid contour-hunting. The integrand function is even, hence it is enough to compute
$$ F(a)=int_{0}^{+infty}frac{sin(ax)}{x(x^2+1)},dx $$
(defined over $text{Re}(a)>0$) at $a=1$. We may notice that
$$ (mathcal{L} F)(s) = int_{0}^{+infty}frac{dx}{(s^2+x^2)(1+x^2)}=frac{pi}{2s(s+1)} $$
hence
$$F(a) = frac{pi}{2}mathcal{L}^{-1}left(frac{1}{s}-frac{1}{s+1}right)(a)=frac{pi}{2}(1-e^{-a}) $$
and
$$ int_{-infty}^{+infty}frac{sin x}{x(x^2+1)},dx = color{blue}{pi(1-e^{-1})}.$$
There is no need to introduce a principal value: the $text{sinc}$ function is continuous and bounded over $mathbb{R}$.
answered Jan 25 at 17:28
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
$begingroup$
Wowww...its good aproach
$endgroup$
– sweety tarika
Jan 25 at 17:50
$begingroup$
Sir,you directly write value of integration...when i try to solve by putting x² =t ...it become lengthy
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
is there any quick method for this integration?
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
@sweetytarika: quicker than the shown one, you mean? You may use Fourier transforms instead of Laplace transforms, it remains a four-liner or so.
$endgroup$
– Jack D'Aurizio
Jan 25 at 18:10
add a comment |
$begingroup$
Let us avoid contour-hunting. The integrand function is even, hence it is enough to compute
$$ F(a)=int_{0}^{+infty}frac{sin(ax)}{x(x^2+1)},dx $$
(defined over $text{Re}(a)>0$) at $a=1$. We may notice that
$$ (mathcal{L} F)(s) = int_{0}^{+infty}frac{dx}{(s^2+x^2)(1+x^2)}=frac{pi}{2s(s+1)} $$
hence
$$F(a) = frac{pi}{2}mathcal{L}^{-1}left(frac{1}{s}-frac{1}{s+1}right)(a)=frac{pi}{2}(1-e^{-a}) $$
and
$$ int_{-infty}^{+infty}frac{sin x}{x(x^2+1)},dx = color{blue}{pi(1-e^{-1})}.$$
There is no need to introduce a principal value: the $text{sinc}$ function is continuous and bounded over $mathbb{R}$.
answered Jan 25 at 17:28
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
Let us avoid contour-hunting. The integrand function is even, hence it is enough to compute
$$ F(a)=int_{0}^{+infty}frac{sin(ax)}{x(x^2+1)},dx $$
(defined over $text{Re}(a)>0$) at $a=1$. We may notice that
$$ (mathcal{L} F)(s) = int_{0}^{+infty}frac{dx}{(s^2+x^2)(1+x^2)}=frac{pi}{2s(s+1)} $$
hence
$$F(a) = frac{pi}{2}mathcal{L}^{-1}left(frac{1}{s}-frac{1}{s+1}right)(a)=frac{pi}{2}(1-e^{-a}) $$
and
$$ int_{-infty}^{+infty}frac{sin x}{x(x^2+1)},dx = color{blue}{pi(1-e^{-1})}.$$
There is no need to introduce a principal value: the $text{sinc}$ function is continuous and bounded over $mathbb{R}$.
answered Jan 25 at 17:28
Jack D'AurizioJack D'Aurizio
291k33284668
answered Jan 25 at 17:28
Jack D'AurizioJack D'Aurizio
291k33284668
answered Jan 25 at 17:28
Jack D'AurizioJack D'Aurizio
291k33284668
answered Jan 25 at 17:28
Jack D'AurizioJack D'Aurizio
291k33284668
291k33284668
$begingroup$
Wowww...its good aproach
$endgroup$
– sweety tarika
Jan 25 at 17:50
$begingroup$
Sir,you directly write value of integration...when i try to solve by putting x² =t ...it become lengthy
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
is there any quick method for this integration?
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
@sweetytarika: quicker than the shown one, you mean? You may use Fourier transforms instead of Laplace transforms, it remains a four-liner or so.
$endgroup$
– Jack D'Aurizio
Jan 25 at 18:10
add a comment |
$begingroup$
Wowww...its good aproach
$endgroup$
– sweety tarika
Jan 25 at 17:50
$begingroup$
Sir,you directly write value of integration...when i try to solve by putting x² =t ...it become lengthy
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
is there any quick method for this integration?
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
@sweetytarika: quicker than the shown one, you mean? You may use Fourier transforms instead of Laplace transforms, it remains a four-liner or so.
$endgroup$
– Jack D'Aurizio
Jan 25 at 18:10
$begingroup$
Wowww...its good aproach
$endgroup$
– sweety tarika
Jan 25 at 17:50
$begingroup$
Wowww...its good aproach
$endgroup$
– sweety tarika
Jan 25 at 17:50
$begingroup$
Sir,you directly write value of integration...when i try to solve by putting x² =t ...it become lengthy
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
Sir,you directly write value of integration...when i try to solve by putting x² =t ...it become lengthy
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
is there any quick method for this integration?
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
is there any quick method for this integration?
$endgroup$
– sweety tarika
Jan 25 at 17:55
$begingroup$
@sweetytarika: quicker than the shown one, you mean? You may use Fourier transforms instead of Laplace transforms, it remains a four-liner or so.
$endgroup$
– Jack D'Aurizio
Jan 25 at 18:10
$begingroup$
@sweetytarika: quicker than the shown one, you mean? You may use Fourier transforms instead of Laplace transforms, it remains a four-liner or so.
$endgroup$
– Jack D'Aurizio
Jan 25 at 18:10
add a comment |
$begingroup$
Here is another approach that like @Jack D'Aurizio approach avoids contour integration altogether.
Let
$$F(a) = int_{-infty}^infty frac{sin (ax)}{x(x^2 + 1)} , dx = 2 int_0^infty frac{sin (ax)}{x(x^2 + 1)} , dx, qquad a > 0.$$
We are required to find $F(1)$.
Using Feynman's trick of differentiating under the integral sign we have
$$F'(a) = 2 int_0^infty frac{cos (ax)}{x^2 + 1} , dx,$$
and
$$F''(a) = - 2 int_0^infty frac{x sin (ax)}{x^2 + 1} , dx.$$
Observe that
$$F''(a) - F(a) = -2 int_0^infty frac{sin (ax)}{x} , dx = -pi.$$
Here the well-known result of
$$int_0^infty frac{sin (ax)}{x} , dx = frac{pi}{2}, qquad a > 0,$$
has been used.
One solving the differential equaltion $F''(a) - F(a) = -pi$ we have
$$F(a) = C_1 e^a + C_2 e^{-a} + pi.$$
The two unknown constants $C_1$ and $C_2$ can be found by noting that $F(0) = 0$ and $F'(0) = pi$. Doing so one finds
$$F(a) = pi(1 - e^{-a}).$$
Finally, setting $a = 1$ one has
$$int_{-infty}^infty frac{sin x}{x(x^2 + 1)} , dx = pi(1 - e^{-1}),$$
as required.
answered Jan 26 at 6:31
omegadotomegadot
6,4072829
$endgroup$
$begingroup$
in begining how you took F(a⁾=2 ,except this step every step is meaningfull
$endgroup$
– sweety tarika
Jan 26 at 11:40
$begingroup$
If you mean the factor of 2 appearing in front of the integral on the first line, that comes from the integrand being an even function between symmetric limits. Is this what you mean?
$endgroup$
– omegadot
Jan 26 at 11:42
$begingroup$
ohkk, got it..thanks
$endgroup$
– sweety tarika
Jan 26 at 11:49
add a comment |
$begingroup$
Here is another approach that like @Jack D'Aurizio approach avoids contour integration altogether.
Let
$$F(a) = int_{-infty}^infty frac{sin (ax)}{x(x^2 + 1)} , dx = 2 int_0^infty frac{sin (ax)}{x(x^2 + 1)} , dx, qquad a > 0.$$
We are required to find $F(1)$.
Using Feynman's trick of differentiating under the integral sign we have
$$F'(a) = 2 int_0^infty frac{cos (ax)}{x^2 + 1} , dx,$$
and
$$F''(a) = - 2 int_0^infty frac{x sin (ax)}{x^2 + 1} , dx.$$
Observe that
$$F''(a) - F(a) = -2 int_0^infty frac{sin (ax)}{x} , dx = -pi.$$
Here the well-known result of
$$int_0^infty frac{sin (ax)}{x} , dx = frac{pi}{2}, qquad a > 0,$$
has been used.
One solving the differential equaltion $F''(a) - F(a) = -pi$ we have
$$F(a) = C_1 e^a + C_2 e^{-a} + pi.$$
The two unknown constants $C_1$ and $C_2$ can be found by noting that $F(0) = 0$ and $F'(0) = pi$. Doing so one finds
$$F(a) = pi(1 - e^{-a}).$$
Finally, setting $a = 1$ one has
$$int_{-infty}^infty frac{sin x}{x(x^2 + 1)} , dx = pi(1 - e^{-1}),$$
as required.
answered Jan 26 at 6:31
omegadotomegadot
6,4072829
$endgroup$
$begingroup$
in begining how you took F(a⁾=2 ,except this step every step is meaningfull
$endgroup$
– sweety tarika
Jan 26 at 11:40
$begingroup$
If you mean the factor of 2 appearing in front of the integral on the first line, that comes from the integrand being an even function between symmetric limits. Is this what you mean?
$endgroup$
– omegadot
Jan 26 at 11:42
$begingroup$
ohkk, got it..thanks
$endgroup$
– sweety tarika
Jan 26 at 11:49
add a comment |
$begingroup$
Here is another approach that like @Jack D'Aurizio approach avoids contour integration altogether.
Let
$$F(a) = int_{-infty}^infty frac{sin (ax)}{x(x^2 + 1)} , dx = 2 int_0^infty frac{sin (ax)}{x(x^2 + 1)} , dx, qquad a > 0.$$
We are required to find $F(1)$.
Using Feynman's trick of differentiating under the integral sign we have
$$F'(a) = 2 int_0^infty frac{cos (ax)}{x^2 + 1} , dx,$$
and
$$F''(a) = - 2 int_0^infty frac{x sin (ax)}{x^2 + 1} , dx.$$
Observe that
$$F''(a) - F(a) = -2 int_0^infty frac{sin (ax)}{x} , dx = -pi.$$
Here the well-known result of
$$int_0^infty frac{sin (ax)}{x} , dx = frac{pi}{2}, qquad a > 0,$$
has been used.
One solving the differential equaltion $F''(a) - F(a) = -pi$ we have
$$F(a) = C_1 e^a + C_2 e^{-a} + pi.$$
The two unknown constants $C_1$ and $C_2$ can be found by noting that $F(0) = 0$ and $F'(0) = pi$. Doing so one finds
$$F(a) = pi(1 - e^{-a}).$$
Finally, setting $a = 1$ one has
$$int_{-infty}^infty frac{sin x}{x(x^2 + 1)} , dx = pi(1 - e^{-1}),$$
as required.
answered Jan 26 at 6:31
omegadotomegadot
6,4072829
$endgroup$
Here is another approach that like @Jack D'Aurizio approach avoids contour integration altogether.
Let
$$F(a) = int_{-infty}^infty frac{sin (ax)}{x(x^2 + 1)} , dx = 2 int_0^infty frac{sin (ax)}{x(x^2 + 1)} , dx, qquad a > 0.$$
We are required to find $F(1)$.
Using Feynman's trick of differentiating under the integral sign we have
$$F'(a) = 2 int_0^infty frac{cos (ax)}{x^2 + 1} , dx,$$
and
$$F''(a) = - 2 int_0^infty frac{x sin (ax)}{x^2 + 1} , dx.$$
Observe that
$$F''(a) - F(a) = -2 int_0^infty frac{sin (ax)}{x} , dx = -pi.$$
Here the well-known result of
$$int_0^infty frac{sin (ax)}{x} , dx = frac{pi}{2}, qquad a > 0,$$
has been used.
One solving the differential equaltion $F''(a) - F(a) = -pi$ we have
$$F(a) = C_1 e^a + C_2 e^{-a} + pi.$$
The two unknown constants $C_1$ and $C_2$ can be found by noting that $F(0) = 0$ and $F'(0) = pi$. Doing so one finds
$$F(a) = pi(1 - e^{-a}).$$
Finally, setting $a = 1$ one has
$$int_{-infty}^infty frac{sin x}{x(x^2 + 1)} , dx = pi(1 - e^{-1}),$$
as required.
answered Jan 26 at 6:31
omegadotomegadot
6,4072829
answered Jan 26 at 6:31
omegadotomegadot
6,4072829
answered Jan 26 at 6:31
omegadotomegadot
6,4072829
answered Jan 26 at 6:31
omegadotomegadot
6,4072829
6,4072829
$begingroup$
in begining how you took F(a⁾=2 ,except this step every step is meaningfull
$endgroup$
– sweety tarika
Jan 26 at 11:40
$begingroup$
If you mean the factor of 2 appearing in front of the integral on the first line, that comes from the integrand being an even function between symmetric limits. Is this what you mean?
$endgroup$
– omegadot
Jan 26 at 11:42
$begingroup$
ohkk, got it..thanks
$endgroup$
– sweety tarika
Jan 26 at 11:49
add a comment |
$begingroup$
in begining how you took F(a⁾=2 ,except this step every step is meaningfull
$endgroup$
– sweety tarika
Jan 26 at 11:40
$begingroup$
If you mean the factor of 2 appearing in front of the integral on the first line, that comes from the integrand being an even function between symmetric limits. Is this what you mean?
$endgroup$
– omegadot
Jan 26 at 11:42
$begingroup$
ohkk, got it..thanks
$endgroup$
– sweety tarika
Jan 26 at 11:49
$begingroup$
in begining how you took F(a⁾=2 ,except this step every step is meaningfull
$endgroup$
– sweety tarika
Jan 26 at 11:40
$begingroup$
in begining how you took F(a⁾=2 ,except this step every step is meaningfull
$endgroup$
– sweety tarika
Jan 26 at 11:40
$begingroup$
If you mean the factor of 2 appearing in front of the integral on the first line, that comes from the integrand being an even function between symmetric limits. Is this what you mean?
$endgroup$
– omegadot
Jan 26 at 11:42
$begingroup$
If you mean the factor of 2 appearing in front of the integral on the first line, that comes from the integrand being an even function between symmetric limits. Is this what you mean?
$endgroup$
– omegadot
Jan 26 at 11:42
$begingroup$
ohkk, got it..thanks
$endgroup$
– sweety tarika
Jan 26 at 11:49
$begingroup$
ohkk, got it..thanks
$endgroup$
– sweety tarika
Jan 26 at 11:49
add a comment |
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$begingroup$
Use Residue theorem. Note that only $0$ and $i$ lie in the upper half plane.
$endgroup$
– Paras Khosla
Jan 25 at 14:35
$begingroup$
You need to show your work, so that you can get direction towards the approach that'll get you to the solution.
$endgroup$
– Paras Khosla
Jan 25 at 14:50
$begingroup$
And when i try to evalute ∫e∧iz/ z(z²⁺1) problem is that z=0 pole comes on boundry of our closed contour
$endgroup$
– sweety tarika
Jan 25 at 15:35
$begingroup$
$z=0$ should not be a problem because in fact $text{Res} bigl(frac{sin z}{z(z^2+1)}, 0bigr)=0$
$endgroup$
– Paras Khosla
Jan 25 at 16:06
$begingroup$
see z=0 is not inside the contour...its on boundry
$endgroup$
– sweety tarika
Jan 25 at 16:35