MHB What is the relationship between the Mellin transform and the sine function?

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The discussion focuses on the Mellin transform of the sine function, defined as the integral of t raised to the power of z minus one multiplied by sin(t). Participants explore the vertical strip in the complex plane where the transform is analytic and discuss methods for finding an analytic representation of the integral involving sin(t). They highlight the use of contour integration as a valid approach for extending results related to the Mellin transform. Additionally, the conversation touches on the behavior of the gamma function in relation to the sine function, noting that the gamma function dominates at large complex values. The relationship between the Mellin transform and sine function is established through these analytical techniques.
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Mellin transform of sine

Define the Mellin transform as

$$\mathcal{MT}\{f(t)\}=\int^\infty_0 t^{z-1} f(t) dt$$ where $$z\in \mathbb{C}$$​
If the transform exists , it is analytic in some vertical strip $$a<\mathrm{Re}(z)<b $$ in the complex plane

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

Additional exercises

1-Find the an analytic function representing

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

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

to prove that

$$\sin(t) = \frac{1}{2 \pi i }\int^{\gamma+i\infty}_{\gamma-i\infty}t^{-z}\mathcal{MT}\{ \sin(t)\} \, dt $$
 
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$ \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.$$ \int_{0}^{\infty} \frac{\sin t}{t^{1-z}} \ dt = \frac{1- \cos t}{t^{1-z}} \Big|_{0}^{\infty} + (1-z) \int_{0}^{\infty} \frac{1- \cos t}{t^{2-z}} \ dt = (1-z) \int_{0}^{\infty} \frac{1- \cos t}{t^{2-z}} \ dt $$

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

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

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

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

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

$$ = \Gamma(z) \frac{2 \sin \left(\frac{\pi z}{2} \right) \cos \left(\frac{\pi z}{2} \right)}{2 \cos \left(\frac{\pi z}{2} \right)} = \Gamma(z) \sin \left( \frac{\pi z}{2} \right)$$
 
Hey RV, Good solution. The third parts isn't that difficult.

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

So we have

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

A third way is by contour integration.
 
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.
 
Well, I have been thinking about that for a long time. It seemed we can extend that result when $$\mathrm{Re}(s)=0 $$ and $$\mathrm{Im}(s) \neq 0$$ . 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

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

is valid for $$\mathrm{Re}(s) >0$$ or $$\mathrm{Re}(s) \leq 0 \, \,\text{ and} \,\,\mathrm{Im}(s) \neq 0 $$
 
$$\frac{1}{2 \pi i } \int_{\gamma - i \infty}^{\gamma + i \infty} \Gamma(z) \sin \left(\frac{\pi z}{2} \right) x^{-z} \ dz = \frac{1}{2 \pi i } \int_{- i \infty}^{ i \infty} \Gamma(z) \sin \left(\frac{\pi z}{2} \right) x^{-z} \ dz$$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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It could be proved using the Stirling estimation that the integral vanishes at large arguments using a half-circle .
 
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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We can use the following approximation for large complex argument

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

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

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