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Mellin transform of sin

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Mellin transform of sine

Define the Mellin transform as

\(\displaystyle \mathcal{MT}\{f(t)\}=\int^\infty_0 t^{z-1} f(t) dt\) where \(\displaystyle z\in \mathbb{C}\)​


If the transform exists , it is analytic in some vertical strip \(\displaystyle a<\mathrm{Re}(z)<b \) in the complex plane

Find the vertical strip of \(\displaystyle \mathcal{MT}\{ \sin(t)\}\)

Additional exercises

1-Find the an analytic function representing

\(\displaystyle \int^\infty_0 t^{z-1} \sin(t) dt\)

2-Use the inverse Mellin transform by integrating along a line (Bromwich integral )

to prove that

\(\displaystyle \sin(t) = \frac{1}{2 \pi i }\int^{\gamma+i\infty}_{\gamma-i\infty}t^{-z}\mathcal{MT}\{ \sin(t)\} \, dt \)
 
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Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$ \displaystyle \int_{0}^{\infty} t^{z-1} \sin t \ dt = \int_{0}^{1} t^{z-1} \sin t \ dt + \int_{1}^{\infty} t^{z-1} \sin t \ dt $


Since $\sin t $ behaves like $t$ near $t=0$, the first integral converges if $\text{Re}(z) > -1 $.

And by Dirichlet's convergence test, the second integral converges if $ \text{Re}(z) < 1$.


So $ \displaystyle \int_{0}^{\infty} t^{z-1} \sin t \ dt$ converges if $-1 < \text{Re}(s) < 1$.


I'm going to integrate by parts first so that when I switch the order of integration it's justified by Tonelli's theorem.


$$ \int_{0}^{\infty} \frac{\sin t}{t^{1-z}} \ dt = \frac{1- \cos t}{t^{1-z}} \Big|_{0}^{\infty} + (1-z) \int_{0}^{\infty} \frac{1- \cos t}{t^{2-z}} \ dt = (1-z) \int_{0}^{\infty} \frac{1- \cos t}{t^{2-z}} \ dt $$

$$ = \frac{1-z}{\Gamma(2-z)} \int_{0}^{\infty} \int_{0}^{\infty} (1 - \cos t) t^{1-z} e^{-tx} \ dx \ dt = \frac{1}{\Gamma(1-z)} \int_{0}^{\infty} \int_{0}^{\infty} (1 - \cos t) t^{1-z} e^{-xt} \ dt \ dx $$

$$= \frac{1}{\Gamma(1-z)} \int_{0}^{\infty} x^{1-z} \left( \frac{1}{x} - \frac{x}{1+x^{2}} \right) \ dx = \frac{1}{\Gamma(1-z)} \int_{0}^{\infty} \frac{x^{-z}}{1+x^{2}} \ dx$$

$$= \frac{1}{\Gamma(1-z)} \int_{0}^{\infty} \frac{x^{(1-z)-1}}{1+x^{2}} \ dx = \frac{1}{\Gamma(1-z)} \frac{\pi}{2} \csc \left( \frac{\pi (1-z)}{2} \right)$$

$$ = \frac{1}{\Gamma (1-z)} \frac{\pi}{2} \sec \left( \frac{\pi z}{2} \right) = \frac{\Gamma (z)}{\Gamma(z) \Gamma(1-z)} \frac{\pi}{2} \sec \left( \frac{\pi z}{2} \right)$$

$$= \frac{\Gamma(z)}{\frac{\pi}{\sin \pi z}} \frac{\pi}{2} \frac{1}{\cos \left(\frac{\pi z}{2}\right)} = \Gamma (z) \frac{\sin \pi z}{2 \cos \left( \frac{\pi z}{2} \right)}$$

$$ = \Gamma(z) \frac{2 \sin \left(\frac{\pi z}{2} \right) \cos \left(\frac{\pi z}{2} \right)}{2 \cos \left(\frac{\pi z}{2} \right)} = \Gamma(z) \sin \left( \frac{\pi z}{2} \right)$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Hey RV, Good solution. The third parts isn't that difficult.

Another way to solve the integral is using the formula
\(\displaystyle \int^\infty_0 t^{z-1} e^{-st} \, dt =\frac{\Gamma(z)}{s^z}\)

So we have

\(\displaystyle \int^\infty_0 t^{z-1} \sin(t) \, dt = \mathcal{Im} \int^\infty_0 t^{z-1} e^{it} \, dt =\mathcal{Im} \frac{\Gamma(z)}{i^z} = \mathcal{Im} \frac{\Gamma(z)}{e^{z \log(i)}} =\Gamma(z) \mathcal{Im}\left( e^{-\frac{\pi}{2} i z } \right)=\Gamma(z)\sin\left( \frac{\pi}{2} \, z \right)\)

A third way is by contour integration.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
The justification for what you did, that is, assuming that formula is valid when $s$ is purely imaginary, comes from contour integration. So contour integration is really the same approach.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Well, I have been thinking about that for a long time. It seemed we can extend that result when \(\displaystyle \mathrm{Re}(s)=0 \) and \(\displaystyle \mathrm{Im}(s) \neq 0\) . The easiest way might be by contours as you said. I failed to find another analytic approach to prove the extended result. But generally it seems if

\(\displaystyle \int^\infty_0 e^{-st} \, t^{z-1} \, dt =\frac{\Gamma(z)}{s^z}\)

is valid for \(\displaystyle \mathrm{Re}(s) >0\) or \(\displaystyle \mathrm{Re}(s) \leq 0 \, \,\text{ and} \,\,\mathrm{Im}(s) \neq 0 \)
 

Random Variable

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MHB Math Helper
Jan 31, 2012
253
$$\frac{1}{2 \pi i } \int_{\gamma - i \infty}^{\gamma + i \infty} \Gamma(z) \sin \left(\frac{\pi z}{2} \right) x^{-z} \ dz = \frac{1}{2 \pi i } \int_{- i \infty}^{ i \infty} \Gamma(z) \sin \left(\frac{\pi z}{2} \right) x^{-z} \ dz$$


Shift the contour to the left.

Then assuming the integral evaluates to zero along the left side of the contour at $-\infty$,

$$ 2 \pi i \int_{- i \infty}^{ i \infty} \Gamma(z) \sin \left(\frac{\pi z}{2} \right) x^{-z} \ dz = \sum_{n=0}^{\infty} \text{Res} \left[ \Gamma (z) \sin \left(\frac{\pi z}{2} \right) x^{-z}, -(2n-1) \right] $$

since the zeros of $\sin \left(\frac{\pi z}{2} \right)$ cancel the simple poles of $\Gamma(z)$ at the negative even integers

$$ = \sum_{n=0}^{\infty} \frac{(-1)^{2n-1}}{(2n-1)!} (-1)^{n} x^{2n-1} = \sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} x^{2n-1} = \sin x $$
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
It could be proved using the Stirling estimation that the integral vanishes at large arguments using a half-circle .
 

Random Variable

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MHB Math Helper
Jan 31, 2012
253
Usually one uses a rectangle because it's easier to understand the behavior of the gamma function on the sides of a rectangle than on a circle.

Basically when $z$ is very, very large in magnitude (and not on the negative real axis), $\Gamma (z)$ is approximately $\left( \frac{z}{e} \right)^z$.

But the issue here isn't the gamma function, it's the sine function.
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
We can use the following approximation for large complex argument

\(\displaystyle |\Gamma(a+ib)| \thicksim \sqrt{2 \pi} |b|^{a-1/2} \, e^{-|b| \pi/2} \,\,\,\,\,\,\, |b| \to \infty\)

I think using this we can see that the gamma dominates the sine function at complex values.