# Mellin transform of sin

#### ZaidAlyafey

##### Well-known member
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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 $$\displaystyle \mathcal{MT}\{ \sin(t)\}$$

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

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MHB Math Helper
$\displaystyle \int_{0}^{\infty} t^{z-1} \sin t \ dt = \int_{0}^{1} t^{z-1} \sin t \ dt + \int_{1}^{\infty} t^{z-1} \sin t \ dt$

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
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

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

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
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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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
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.