Mixing
Mellin transform of $sin x$ aka $int^{infty}_0 x^{s-1}sin x dx $
$begingroup$
I am trying to find the Mellin transform of $sin x $, put in other words to solve:
$$int^{infty}_0 x^{s-1}sin x mathrm{d} x $$
And I know that the answer is:
$$Gamma(s) sin left(frac{pi s}{2}right)$$
From several tables on the internet but I couldn't find any justification.
How can this identity be proven?
calculus integration complex-analysis contour-integration mellin-transform
edited Jan 26 at 20:35
Zacky
7,89511061
asked May 5 '13 at 18:38
Max CliffordMax Clifford
524
$endgroup$
add a comment |
$begingroup$
I am trying to find the Mellin transform of $sin x $, put in other words to solve:
$$int^{infty}_0 x^{s-1}sin x mathrm{d} x $$
And I know that the answer is:
$$Gamma(s) sin left(frac{pi s}{2}right)$$
From several tables on the internet but I couldn't find any justification.
How can this identity be proven?
calculus integration complex-analysis contour-integration mellin-transform
edited Jan 26 at 20:35
Zacky
7,89511061
asked May 5 '13 at 18:38
Max CliffordMax Clifford
524
$endgroup$
add a comment |
$begingroup$
I am trying to find the Mellin transform of $sin x $, put in other words to solve:
$$int^{infty}_0 x^{s-1}sin x mathrm{d} x $$
And I know that the answer is:
$$Gamma(s) sin left(frac{pi s}{2}right)$$
From several tables on the internet but I couldn't find any justification.
How can this identity be proven?
calculus integration complex-analysis contour-integration mellin-transform
edited Jan 26 at 20:35
Zacky
7,89511061
asked May 5 '13 at 18:38
Max CliffordMax Clifford
524
$endgroup$
I am trying to find the Mellin transform of $sin x $, put in other words to solve:
$$int^{infty}_0 x^{s-1}sin x mathrm{d} x $$
And I know that the answer is:
$$Gamma(s) sin left(frac{pi s}{2}right)$$
From several tables on the internet but I couldn't find any justification.
How can this identity be proven?
calculus integration complex-analysis contour-integration mellin-transform
calculus integration complex-analysis contour-integration mellin-transform
edited Jan 26 at 20:35
Zacky
7,89511061
asked May 5 '13 at 18:38
Max CliffordMax Clifford
524
edited Jan 26 at 20:35
Zacky
7,89511061
asked May 5 '13 at 18:38
Max CliffordMax Clifford
524
edited Jan 26 at 20:35
Zacky
7,89511061
edited Jan 26 at 20:35
Zacky
7,89511061
edited Jan 26 at 20:35
Zacky
7,89511061
7,89511061
asked May 5 '13 at 18:38
Max CliffordMax Clifford
524
asked May 5 '13 at 18:38
Max CliffordMax Clifford
524
asked May 5 '13 at 18:38
Max CliffordMax Clifford
524
524
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The basic idea is to use Euler's formula: $sin x = dfrac{e^{ix}-e^{-ix}}{2i}$.
Let's look at the Mellin transform of $e^{ix}$:
$$int_0^infty e^{ix} x^{s-1},dx = begin{bmatrix} x=it \ dx = i,dtend{bmatrix} = int_{0}^{-icdotinfty} e^{-t} (it)^{s-1} i,dt = -i^s int_{-icdot infty}^0 e^{-t}t^{s-1},dt.$$
For certain values of $s$ (I will leave it to you to work out the details), you can deform the integral over the negative imaginary axis to an integral over the positive real axis:
$$int_{-icdot infty}^0 e^{-t}t^{s-1},dt = -int_0^infty e^{-t}t^{s-1},dt = -Gamma(s).$$
(Add a large quarter circle in the fourth quadrant, use Cauchys integral theorem and estimate $f(z) = e^{-z}z^{s-1}$ on the new quarter circle.)
Summing up, we get that Mellin transform of $e^{ix}$ is
$$i^sGamma(s) = exp(ipi s/2) Gamma(s).$$
Similarly, the Mellin transform of $e^{-ix}$ turns out to be $$i^{-s}Gamma(s) = exp(-ipi s/2) Gamma(s).$$
Forming the appropriate linear combination, the Mellin transform of $sin x$ ends up as
$$ frac{exp(ipi s/2) + exp(-ipi s/2)}{2i} Gamma(s) = sinfrac{pi s}2 Gamma(s).$$
answered May 6 '13 at 9:25
mrfmrf
37.6k64786
$endgroup$
$begingroup$
Thanks mrf for puting so much time into that, I really appreciate it. The step I'm looking for is the the bit you've put in brackets about the fourth quadrant. How do you know to use that quadrant and not another? And how may I deform the curve? Thanks
$endgroup$
– Max Clifford
May 6 '13 at 19:26
$begingroup$
@MaxClifford You want to integrate along the ray from $-iinfty$ to $0$, which really means integrating from $-iR$ to $0$ and letting $Rtoinfty$. You want to relate your integral with one over the positive real axis (i.e. from $0$ to $R$), to be able to compare with the $Gamma$-function. Draw a picture. One of the few reasonable ways to connect the two rays is with a quarter circle in the fourth quadrant. (I hope you're familiar with ideas from complex analysis, in particular the residue theorem.)
$endgroup$
– mrf
May 6 '13 at 19:42
add a comment |
$begingroup$
Alternatively, the Mellin transform for $sin x$ can be found by employing the following useful property for the Laplace transform:
$$int_0^infty f(x) g(x) , dx = int_0^infty mathcal{L} {f(x)} (t) cdot mathcal{L}^{-1} {g(x)} (t) , dt.$$
Noting that
$$mathcal{L} {sin x}(t) = frac{1}{1 + t^2},$$
and
$$mathcal{L}^{-1} left {frac{1}{x^{1-s}} right } (t)= frac{1}{Gamma (1 - s)} mathcal{L}^{-1} left {frac{Gamma (1 - s)}{x^{1-s}} right } (t) = frac{t^{-s}}{Gamma (1 - s)},$$
then
begin{align}
mathcal{M} {sin x} &= int_0^infty sin x cdot frac{1}{x^{1 - s}} , dx\
&= int_0^infty mathcal{L} {sin x} (t) cdot mathcal{L}^{-1} left {frac{1}{x^{1 - s}} right } (t) , dt\
&= frac{1}{Gamma (1 - s)} int_0^infty frac{t^{-s}}{1 + t^2} , dt.
end{align}
Setting $u = t^2$, one has
begin{align}
mathcal{M} {sin x} &= frac{1}{2 Gamma (1 - s)} int_0^infty frac{u^{-frac{s}{2} - frac{1}{2}}}{1 + u} , du\
&= frac{1}{2 Gamma (1 - s)} operatorname{B} left (frac{1}{2} - frac{s}{2}, frac{1}{2} + frac{s}{2} right ) tag1\
&= frac{1}{2 Gamma (1 - s)} Gamma left (frac{1}{2} - frac{s}{2} right ) Gamma left (frac{1}{2} + frac{s}{2} right ) tag2\
&= frac{1}{2 Gamma (1 - s)} Gamma left [1 - left (frac{1}{2} + frac{s}{2} right ) right ] Gamma left (frac{1}{2} + frac{s}{2} right )\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{sin left (frac{pi}{2} + frac{pi s}{2} right )} tag3\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{cos left (frac{pi s}{2} right )}\
&= frac{Gamma (s) sin (pi s)}{2 pi} cdot frac{pi}{cos left (frac{pi s}{2} right )} tag4\
&=frac{Gamma (s) sin left (frac{pi s}{2} right ) cos left (frac{pi s}{2} right )}{cos left (frac{pi s}{2} right )}\
&= Gamma (s) sin left (frac{pi s}{2} right )
end{align}
This is valid for $-1 < s < 1$.
Explanation
(1) Using $operatorname{B} (x,y) = displaystyle{int_0^infty frac{t^{x - 1}}{(1 + t)^{x + y}} , dt}$.
(2) Using $operatorname{B}(x,y) = dfrac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(3) Using the reflexion formula for the gamma function: $Gamma (1 - z) Gamma (z) = dfrac{pi}{sin (pi z)}$.
(4) Again using the reflexion formula for the gamma function.
answered Feb 6 at 1:19
omegadotomegadot
6,4492829
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add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
The basic idea is to use Euler's formula: $sin x = dfrac{e^{ix}-e^{-ix}}{2i}$.
Let's look at the Mellin transform of $e^{ix}$:
$$int_0^infty e^{ix} x^{s-1},dx = begin{bmatrix} x=it \ dx = i,dtend{bmatrix} = int_{0}^{-icdotinfty} e^{-t} (it)^{s-1} i,dt = -i^s int_{-icdot infty}^0 e^{-t}t^{s-1},dt.$$
For certain values of $s$ (I will leave it to you to work out the details), you can deform the integral over the negative imaginary axis to an integral over the positive real axis:
$$int_{-icdot infty}^0 e^{-t}t^{s-1},dt = -int_0^infty e^{-t}t^{s-1},dt = -Gamma(s).$$
(Add a large quarter circle in the fourth quadrant, use Cauchys integral theorem and estimate $f(z) = e^{-z}z^{s-1}$ on the new quarter circle.)
Summing up, we get that Mellin transform of $e^{ix}$ is
$$i^sGamma(s) = exp(ipi s/2) Gamma(s).$$
Similarly, the Mellin transform of $e^{-ix}$ turns out to be $$i^{-s}Gamma(s) = exp(-ipi s/2) Gamma(s).$$
Forming the appropriate linear combination, the Mellin transform of $sin x$ ends up as
$$ frac{exp(ipi s/2) + exp(-ipi s/2)}{2i} Gamma(s) = sinfrac{pi s}2 Gamma(s).$$
answered May 6 '13 at 9:25
mrfmrf
37.6k64786
$endgroup$
$begingroup$
Thanks mrf for puting so much time into that, I really appreciate it. The step I'm looking for is the the bit you've put in brackets about the fourth quadrant. How do you know to use that quadrant and not another? And how may I deform the curve? Thanks
$endgroup$
– Max Clifford
May 6 '13 at 19:26
$begingroup$
@MaxClifford You want to integrate along the ray from $-iinfty$ to $0$, which really means integrating from $-iR$ to $0$ and letting $Rtoinfty$. You want to relate your integral with one over the positive real axis (i.e. from $0$ to $R$), to be able to compare with the $Gamma$-function. Draw a picture. One of the few reasonable ways to connect the two rays is with a quarter circle in the fourth quadrant. (I hope you're familiar with ideas from complex analysis, in particular the residue theorem.)
$endgroup$
– mrf
May 6 '13 at 19:42
add a comment |
$begingroup$
The basic idea is to use Euler's formula: $sin x = dfrac{e^{ix}-e^{-ix}}{2i}$.
Let's look at the Mellin transform of $e^{ix}$:
$$int_0^infty e^{ix} x^{s-1},dx = begin{bmatrix} x=it \ dx = i,dtend{bmatrix} = int_{0}^{-icdotinfty} e^{-t} (it)^{s-1} i,dt = -i^s int_{-icdot infty}^0 e^{-t}t^{s-1},dt.$$
For certain values of $s$ (I will leave it to you to work out the details), you can deform the integral over the negative imaginary axis to an integral over the positive real axis:
$$int_{-icdot infty}^0 e^{-t}t^{s-1},dt = -int_0^infty e^{-t}t^{s-1},dt = -Gamma(s).$$
(Add a large quarter circle in the fourth quadrant, use Cauchys integral theorem and estimate $f(z) = e^{-z}z^{s-1}$ on the new quarter circle.)
Summing up, we get that Mellin transform of $e^{ix}$ is
$$i^sGamma(s) = exp(ipi s/2) Gamma(s).$$
Similarly, the Mellin transform of $e^{-ix}$ turns out to be $$i^{-s}Gamma(s) = exp(-ipi s/2) Gamma(s).$$
Forming the appropriate linear combination, the Mellin transform of $sin x$ ends up as
$$ frac{exp(ipi s/2) + exp(-ipi s/2)}{2i} Gamma(s) = sinfrac{pi s}2 Gamma(s).$$
answered May 6 '13 at 9:25
mrfmrf
37.6k64786
$endgroup$
$begingroup$
Thanks mrf for puting so much time into that, I really appreciate it. The step I'm looking for is the the bit you've put in brackets about the fourth quadrant. How do you know to use that quadrant and not another? And how may I deform the curve? Thanks
$endgroup$
– Max Clifford
May 6 '13 at 19:26
$begingroup$
@MaxClifford You want to integrate along the ray from $-iinfty$ to $0$, which really means integrating from $-iR$ to $0$ and letting $Rtoinfty$. You want to relate your integral with one over the positive real axis (i.e. from $0$ to $R$), to be able to compare with the $Gamma$-function. Draw a picture. One of the few reasonable ways to connect the two rays is with a quarter circle in the fourth quadrant. (I hope you're familiar with ideas from complex analysis, in particular the residue theorem.)
$endgroup$
– mrf
May 6 '13 at 19:42
add a comment |
$begingroup$
The basic idea is to use Euler's formula: $sin x = dfrac{e^{ix}-e^{-ix}}{2i}$.
Let's look at the Mellin transform of $e^{ix}$:
$$int_0^infty e^{ix} x^{s-1},dx = begin{bmatrix} x=it \ dx = i,dtend{bmatrix} = int_{0}^{-icdotinfty} e^{-t} (it)^{s-1} i,dt = -i^s int_{-icdot infty}^0 e^{-t}t^{s-1},dt.$$
For certain values of $s$ (I will leave it to you to work out the details), you can deform the integral over the negative imaginary axis to an integral over the positive real axis:
$$int_{-icdot infty}^0 e^{-t}t^{s-1},dt = -int_0^infty e^{-t}t^{s-1},dt = -Gamma(s).$$
(Add a large quarter circle in the fourth quadrant, use Cauchys integral theorem and estimate $f(z) = e^{-z}z^{s-1}$ on the new quarter circle.)
Summing up, we get that Mellin transform of $e^{ix}$ is
$$i^sGamma(s) = exp(ipi s/2) Gamma(s).$$
Similarly, the Mellin transform of $e^{-ix}$ turns out to be $$i^{-s}Gamma(s) = exp(-ipi s/2) Gamma(s).$$
Forming the appropriate linear combination, the Mellin transform of $sin x$ ends up as
$$ frac{exp(ipi s/2) + exp(-ipi s/2)}{2i} Gamma(s) = sinfrac{pi s}2 Gamma(s).$$
answered May 6 '13 at 9:25
mrfmrf
37.6k64786
$endgroup$
The basic idea is to use Euler's formula: $sin x = dfrac{e^{ix}-e^{-ix}}{2i}$.
Let's look at the Mellin transform of $e^{ix}$:
$$int_0^infty e^{ix} x^{s-1},dx = begin{bmatrix} x=it \ dx = i,dtend{bmatrix} = int_{0}^{-icdotinfty} e^{-t} (it)^{s-1} i,dt = -i^s int_{-icdot infty}^0 e^{-t}t^{s-1},dt.$$
For certain values of $s$ (I will leave it to you to work out the details), you can deform the integral over the negative imaginary axis to an integral over the positive real axis:
$$int_{-icdot infty}^0 e^{-t}t^{s-1},dt = -int_0^infty e^{-t}t^{s-1},dt = -Gamma(s).$$
(Add a large quarter circle in the fourth quadrant, use Cauchys integral theorem and estimate $f(z) = e^{-z}z^{s-1}$ on the new quarter circle.)
Summing up, we get that Mellin transform of $e^{ix}$ is
$$i^sGamma(s) = exp(ipi s/2) Gamma(s).$$
Similarly, the Mellin transform of $e^{-ix}$ turns out to be $$i^{-s}Gamma(s) = exp(-ipi s/2) Gamma(s).$$
Forming the appropriate linear combination, the Mellin transform of $sin x$ ends up as
$$ frac{exp(ipi s/2) + exp(-ipi s/2)}{2i} Gamma(s) = sinfrac{pi s}2 Gamma(s).$$
answered May 6 '13 at 9:25
mrfmrf
37.6k64786
edited May 6 '13 at 9:38
answered May 6 '13 at 9:25
mrfmrf
37.6k64786
answered May 6 '13 at 9:25
mrfmrf
37.6k64786
answered May 6 '13 at 9:25
mrfmrf
37.6k64786
37.6k64786
$begingroup$
Thanks mrf for puting so much time into that, I really appreciate it. The step I'm looking for is the the bit you've put in brackets about the fourth quadrant. How do you know to use that quadrant and not another? And how may I deform the curve? Thanks
$endgroup$
– Max Clifford
May 6 '13 at 19:26
$begingroup$
@MaxClifford You want to integrate along the ray from $-iinfty$ to $0$, which really means integrating from $-iR$ to $0$ and letting $Rtoinfty$. You want to relate your integral with one over the positive real axis (i.e. from $0$ to $R$), to be able to compare with the $Gamma$-function. Draw a picture. One of the few reasonable ways to connect the two rays is with a quarter circle in the fourth quadrant. (I hope you're familiar with ideas from complex analysis, in particular the residue theorem.)
$endgroup$
– mrf
May 6 '13 at 19:42
add a comment |
$begingroup$
Thanks mrf for puting so much time into that, I really appreciate it. The step I'm looking for is the the bit you've put in brackets about the fourth quadrant. How do you know to use that quadrant and not another? And how may I deform the curve? Thanks
$endgroup$
– Max Clifford
May 6 '13 at 19:26
$begingroup$
@MaxClifford You want to integrate along the ray from $-iinfty$ to $0$, which really means integrating from $-iR$ to $0$ and letting $Rtoinfty$. You want to relate your integral with one over the positive real axis (i.e. from $0$ to $R$), to be able to compare with the $Gamma$-function. Draw a picture. One of the few reasonable ways to connect the two rays is with a quarter circle in the fourth quadrant. (I hope you're familiar with ideas from complex analysis, in particular the residue theorem.)
$endgroup$
– mrf
May 6 '13 at 19:42
$begingroup$
Thanks mrf for puting so much time into that, I really appreciate it. The step I'm looking for is the the bit you've put in brackets about the fourth quadrant. How do you know to use that quadrant and not another? And how may I deform the curve? Thanks
$endgroup$
– Max Clifford
May 6 '13 at 19:26
$begingroup$
Thanks mrf for puting so much time into that, I really appreciate it. The step I'm looking for is the the bit you've put in brackets about the fourth quadrant. How do you know to use that quadrant and not another? And how may I deform the curve? Thanks
$endgroup$
– Max Clifford
May 6 '13 at 19:26
$begingroup$
@MaxClifford You want to integrate along the ray from $-iinfty$ to $0$, which really means integrating from $-iR$ to $0$ and letting $Rtoinfty$. You want to relate your integral with one over the positive real axis (i.e. from $0$ to $R$), to be able to compare with the $Gamma$-function. Draw a picture. One of the few reasonable ways to connect the two rays is with a quarter circle in the fourth quadrant. (I hope you're familiar with ideas from complex analysis, in particular the residue theorem.)
$endgroup$
– mrf
May 6 '13 at 19:42
$begingroup$
@MaxClifford You want to integrate along the ray from $-iinfty$ to $0$, which really means integrating from $-iR$ to $0$ and letting $Rtoinfty$. You want to relate your integral with one over the positive real axis (i.e. from $0$ to $R$), to be able to compare with the $Gamma$-function. Draw a picture. One of the few reasonable ways to connect the two rays is with a quarter circle in the fourth quadrant. (I hope you're familiar with ideas from complex analysis, in particular the residue theorem.)
$endgroup$
– mrf
May 6 '13 at 19:42
add a comment |
$begingroup$
Alternatively, the Mellin transform for $sin x$ can be found by employing the following useful property for the Laplace transform:
$$int_0^infty f(x) g(x) , dx = int_0^infty mathcal{L} {f(x)} (t) cdot mathcal{L}^{-1} {g(x)} (t) , dt.$$
Noting that
$$mathcal{L} {sin x}(t) = frac{1}{1 + t^2},$$
and
$$mathcal{L}^{-1} left {frac{1}{x^{1-s}} right } (t)= frac{1}{Gamma (1 - s)} mathcal{L}^{-1} left {frac{Gamma (1 - s)}{x^{1-s}} right } (t) = frac{t^{-s}}{Gamma (1 - s)},$$
then
begin{align}
mathcal{M} {sin x} &= int_0^infty sin x cdot frac{1}{x^{1 - s}} , dx\
&= int_0^infty mathcal{L} {sin x} (t) cdot mathcal{L}^{-1} left {frac{1}{x^{1 - s}} right } (t) , dt\
&= frac{1}{Gamma (1 - s)} int_0^infty frac{t^{-s}}{1 + t^2} , dt.
end{align}
Setting $u = t^2$, one has
begin{align}
mathcal{M} {sin x} &= frac{1}{2 Gamma (1 - s)} int_0^infty frac{u^{-frac{s}{2} - frac{1}{2}}}{1 + u} , du\
&= frac{1}{2 Gamma (1 - s)} operatorname{B} left (frac{1}{2} - frac{s}{2}, frac{1}{2} + frac{s}{2} right ) tag1\
&= frac{1}{2 Gamma (1 - s)} Gamma left (frac{1}{2} - frac{s}{2} right ) Gamma left (frac{1}{2} + frac{s}{2} right ) tag2\
&= frac{1}{2 Gamma (1 - s)} Gamma left [1 - left (frac{1}{2} + frac{s}{2} right ) right ] Gamma left (frac{1}{2} + frac{s}{2} right )\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{sin left (frac{pi}{2} + frac{pi s}{2} right )} tag3\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{cos left (frac{pi s}{2} right )}\
&= frac{Gamma (s) sin (pi s)}{2 pi} cdot frac{pi}{cos left (frac{pi s}{2} right )} tag4\
&=frac{Gamma (s) sin left (frac{pi s}{2} right ) cos left (frac{pi s}{2} right )}{cos left (frac{pi s}{2} right )}\
&= Gamma (s) sin left (frac{pi s}{2} right )
end{align}
This is valid for $-1 < s < 1$.
Explanation
(1) Using $operatorname{B} (x,y) = displaystyle{int_0^infty frac{t^{x - 1}}{(1 + t)^{x + y}} , dt}$.
(2) Using $operatorname{B}(x,y) = dfrac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(3) Using the reflexion formula for the gamma function: $Gamma (1 - z) Gamma (z) = dfrac{pi}{sin (pi z)}$.
(4) Again using the reflexion formula for the gamma function.
answered Feb 6 at 1:19
omegadotomegadot
6,4492829
$endgroup$
add a comment |
$begingroup$
Alternatively, the Mellin transform for $sin x$ can be found by employing the following useful property for the Laplace transform:
$$int_0^infty f(x) g(x) , dx = int_0^infty mathcal{L} {f(x)} (t) cdot mathcal{L}^{-1} {g(x)} (t) , dt.$$
Noting that
$$mathcal{L} {sin x}(t) = frac{1}{1 + t^2},$$
and
$$mathcal{L}^{-1} left {frac{1}{x^{1-s}} right } (t)= frac{1}{Gamma (1 - s)} mathcal{L}^{-1} left {frac{Gamma (1 - s)}{x^{1-s}} right } (t) = frac{t^{-s}}{Gamma (1 - s)},$$
then
begin{align}
mathcal{M} {sin x} &= int_0^infty sin x cdot frac{1}{x^{1 - s}} , dx\
&= int_0^infty mathcal{L} {sin x} (t) cdot mathcal{L}^{-1} left {frac{1}{x^{1 - s}} right } (t) , dt\
&= frac{1}{Gamma (1 - s)} int_0^infty frac{t^{-s}}{1 + t^2} , dt.
end{align}
Setting $u = t^2$, one has
begin{align}
mathcal{M} {sin x} &= frac{1}{2 Gamma (1 - s)} int_0^infty frac{u^{-frac{s}{2} - frac{1}{2}}}{1 + u} , du\
&= frac{1}{2 Gamma (1 - s)} operatorname{B} left (frac{1}{2} - frac{s}{2}, frac{1}{2} + frac{s}{2} right ) tag1\
&= frac{1}{2 Gamma (1 - s)} Gamma left (frac{1}{2} - frac{s}{2} right ) Gamma left (frac{1}{2} + frac{s}{2} right ) tag2\
&= frac{1}{2 Gamma (1 - s)} Gamma left [1 - left (frac{1}{2} + frac{s}{2} right ) right ] Gamma left (frac{1}{2} + frac{s}{2} right )\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{sin left (frac{pi}{2} + frac{pi s}{2} right )} tag3\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{cos left (frac{pi s}{2} right )}\
&= frac{Gamma (s) sin (pi s)}{2 pi} cdot frac{pi}{cos left (frac{pi s}{2} right )} tag4\
&=frac{Gamma (s) sin left (frac{pi s}{2} right ) cos left (frac{pi s}{2} right )}{cos left (frac{pi s}{2} right )}\
&= Gamma (s) sin left (frac{pi s}{2} right )
end{align}
This is valid for $-1 < s < 1$.
Explanation
(1) Using $operatorname{B} (x,y) = displaystyle{int_0^infty frac{t^{x - 1}}{(1 + t)^{x + y}} , dt}$.
(2) Using $operatorname{B}(x,y) = dfrac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(3) Using the reflexion formula for the gamma function: $Gamma (1 - z) Gamma (z) = dfrac{pi}{sin (pi z)}$.
(4) Again using the reflexion formula for the gamma function.
answered Feb 6 at 1:19
omegadotomegadot
6,4492829
$endgroup$
add a comment |
$begingroup$
Alternatively, the Mellin transform for $sin x$ can be found by employing the following useful property for the Laplace transform:
$$int_0^infty f(x) g(x) , dx = int_0^infty mathcal{L} {f(x)} (t) cdot mathcal{L}^{-1} {g(x)} (t) , dt.$$
Noting that
$$mathcal{L} {sin x}(t) = frac{1}{1 + t^2},$$
and
$$mathcal{L}^{-1} left {frac{1}{x^{1-s}} right } (t)= frac{1}{Gamma (1 - s)} mathcal{L}^{-1} left {frac{Gamma (1 - s)}{x^{1-s}} right } (t) = frac{t^{-s}}{Gamma (1 - s)},$$
then
begin{align}
mathcal{M} {sin x} &= int_0^infty sin x cdot frac{1}{x^{1 - s}} , dx\
&= int_0^infty mathcal{L} {sin x} (t) cdot mathcal{L}^{-1} left {frac{1}{x^{1 - s}} right } (t) , dt\
&= frac{1}{Gamma (1 - s)} int_0^infty frac{t^{-s}}{1 + t^2} , dt.
end{align}
Setting $u = t^2$, one has
begin{align}
mathcal{M} {sin x} &= frac{1}{2 Gamma (1 - s)} int_0^infty frac{u^{-frac{s}{2} - frac{1}{2}}}{1 + u} , du\
&= frac{1}{2 Gamma (1 - s)} operatorname{B} left (frac{1}{2} - frac{s}{2}, frac{1}{2} + frac{s}{2} right ) tag1\
&= frac{1}{2 Gamma (1 - s)} Gamma left (frac{1}{2} - frac{s}{2} right ) Gamma left (frac{1}{2} + frac{s}{2} right ) tag2\
&= frac{1}{2 Gamma (1 - s)} Gamma left [1 - left (frac{1}{2} + frac{s}{2} right ) right ] Gamma left (frac{1}{2} + frac{s}{2} right )\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{sin left (frac{pi}{2} + frac{pi s}{2} right )} tag3\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{cos left (frac{pi s}{2} right )}\
&= frac{Gamma (s) sin (pi s)}{2 pi} cdot frac{pi}{cos left (frac{pi s}{2} right )} tag4\
&=frac{Gamma (s) sin left (frac{pi s}{2} right ) cos left (frac{pi s}{2} right )}{cos left (frac{pi s}{2} right )}\
&= Gamma (s) sin left (frac{pi s}{2} right )
end{align}
This is valid for $-1 < s < 1$.
Explanation
(1) Using $operatorname{B} (x,y) = displaystyle{int_0^infty frac{t^{x - 1}}{(1 + t)^{x + y}} , dt}$.
(2) Using $operatorname{B}(x,y) = dfrac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(3) Using the reflexion formula for the gamma function: $Gamma (1 - z) Gamma (z) = dfrac{pi}{sin (pi z)}$.
(4) Again using the reflexion formula for the gamma function.
answered Feb 6 at 1:19
omegadotomegadot
6,4492829
$endgroup$
Alternatively, the Mellin transform for $sin x$ can be found by employing the following useful property for the Laplace transform:
$$int_0^infty f(x) g(x) , dx = int_0^infty mathcal{L} {f(x)} (t) cdot mathcal{L}^{-1} {g(x)} (t) , dt.$$
Noting that
$$mathcal{L} {sin x}(t) = frac{1}{1 + t^2},$$
and
$$mathcal{L}^{-1} left {frac{1}{x^{1-s}} right } (t)= frac{1}{Gamma (1 - s)} mathcal{L}^{-1} left {frac{Gamma (1 - s)}{x^{1-s}} right } (t) = frac{t^{-s}}{Gamma (1 - s)},$$
then
begin{align}
mathcal{M} {sin x} &= int_0^infty sin x cdot frac{1}{x^{1 - s}} , dx\
&= int_0^infty mathcal{L} {sin x} (t) cdot mathcal{L}^{-1} left {frac{1}{x^{1 - s}} right } (t) , dt\
&= frac{1}{Gamma (1 - s)} int_0^infty frac{t^{-s}}{1 + t^2} , dt.
end{align}
Setting $u = t^2$, one has
begin{align}
mathcal{M} {sin x} &= frac{1}{2 Gamma (1 - s)} int_0^infty frac{u^{-frac{s}{2} - frac{1}{2}}}{1 + u} , du\
&= frac{1}{2 Gamma (1 - s)} operatorname{B} left (frac{1}{2} - frac{s}{2}, frac{1}{2} + frac{s}{2} right ) tag1\
&= frac{1}{2 Gamma (1 - s)} Gamma left (frac{1}{2} - frac{s}{2} right ) Gamma left (frac{1}{2} + frac{s}{2} right ) tag2\
&= frac{1}{2 Gamma (1 - s)} Gamma left [1 - left (frac{1}{2} + frac{s}{2} right ) right ] Gamma left (frac{1}{2} + frac{s}{2} right )\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{sin left (frac{pi}{2} + frac{pi s}{2} right )} tag3\
&= frac{1}{2 Gamma (1 - s)} frac{pi}{cos left (frac{pi s}{2} right )}\
&= frac{Gamma (s) sin (pi s)}{2 pi} cdot frac{pi}{cos left (frac{pi s}{2} right )} tag4\
&=frac{Gamma (s) sin left (frac{pi s}{2} right ) cos left (frac{pi s}{2} right )}{cos left (frac{pi s}{2} right )}\
&= Gamma (s) sin left (frac{pi s}{2} right )
end{align}
This is valid for $-1 < s < 1$.
Explanation
(1) Using $operatorname{B} (x,y) = displaystyle{int_0^infty frac{t^{x - 1}}{(1 + t)^{x + y}} , dt}$.
(2) Using $operatorname{B}(x,y) = dfrac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(3) Using the reflexion formula for the gamma function: $Gamma (1 - z) Gamma (z) = dfrac{pi}{sin (pi z)}$.
(4) Again using the reflexion formula for the gamma function.
answered Feb 6 at 1:19
omegadotomegadot
6,4492829
edited Feb 6 at 1:32
answered Feb 6 at 1:19
omegadotomegadot
6,4492829
answered Feb 6 at 1:19
omegadotomegadot
6,4492829
answered Feb 6 at 1:19
omegadotomegadot
6,4492829
6,4492829
add a comment |
add a comment |
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