Is $\int\limits_0^\infty\frac{\sin y}{y^{s+1}}dy=-\Gamma(-s)\sin(\frac{\pi s}{2})$ for $\operatorname{Re}(s)\in (-1,0)$ obvious?

I realised this question has been asked before as you can see here. Anyway I will write down my solution here again. First of all consider Ramanuajan's Master Theorem.

Ramanujan's Master Theorem

Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form $$f(x)=\sum_{k=0}^{\infty}\frac{\phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by $$\int_0^{\infty}x^{p-1}f(x)dx=\Gamma(p)\phi(-p)$$

In order to use this Theorem we may expand the sine function as a series followed by the substitution $y^2=t$ which yields to

$$\begin{align*} \mathfrak{I}=\int_0^{\infty}y^{-s-1}\sin(y)dy&=\int_0^{\infty}y^{-s-1}\sum_{n=0}^\infty (-1)^n \frac{y^{2n+1}}{(2n+1)!}dy\\ &=\frac12\int_0^{\infty}y^{-s-1}\sum_{n=0}^\infty (-1)^n \frac{n!/(2n+1)!}{n!}(-y^2)^n[2ydy]\\ &=\frac12\int_0^\infty t^{-(s+1)/2}\sum_{n=0}^\infty\frac{n!/(2n+1)!}{n!}(-t)^ndt \end{align*}$$

Now we can use Ramanuajan's Master Theorem by setting $p=-\frac{s-1}2$ and $\phi(n)=\frac{n!}{(2n+1)!}=\frac{\Gamma(n+1)}{\Gamma(2(n+1))}$ and so we get

$$\begin{align*} \mathfrak{I}=\frac12\int_0^\infty t^{-(s+1)/2}\sum_{n=0}^\infty\frac{n!/(2n+1)!}{n!}(-t)^ndt&=\frac12\Gamma\left(-\frac{s-1}2\right)\frac{\Gamma\left(1+\frac{s-1}2\right)}{\Gamma\left(2\left(\frac{s-1}2+1\right)\right)}\\ &=\frac1{2\Gamma(s+1)}\Gamma\left(\frac{s+1}2\right)\Gamma\left(-\frac{s-1}2\right)\tag1\\ &=\frac1{2\Gamma(s+1)}\frac{\pi}{\sin\left(\pi\frac{s+1}2\right)}\\ &=\frac1{2\Gamma(s+1)}\frac{\pi}{\cos\left(\frac{\pi s}2\right)}\\ &=\frac{\pi}{\Gamma(s+1)}\frac{\sin\left(\frac{\pi s}2\right)}{2\sin\left(\frac{\pi s}2\right)\cos\left(\frac{\pi s}2\right)}\\ &=-\sin\left(\frac{\pi s}2\right)\frac{\pi}{\Gamma(s+1)\sin(\pi(s+1))}\tag2\\ &=-\sin\left(\frac{\pi s}2\right)\Gamma(-s) \end{align*}$$

$$\therefore~\mathfrak{I}=\int_0^{\infty}y^{-s-1}\sin(y)dy~=~-\Gamma(-s)\sin\left(\frac{\pi s}2\right)$$

For the simplification of the final solution we excessively used Euler's Reflection Formula which is a key property of the Gamma Function. Within line $(1)$ we applied the formula for $z=\frac{s+1}2$ and within line $(2)$ for $z=s+1$. The trigonometric reshaping utilized the double-angle formula as well as the periodic property of the sine function.


I'd say most of the book relies heavily on the same kind of derivation : complex analysis, change of variable, change of contour, recognizing famous integrals, restricting to domains where everything is easier then extending by continuity/analyticity.

For $\Re(s) < 0$ and $\Re(e^a) >0$ then $$\int_0^\infty t^{-s-1} e^{-e^a t}dt = \int_0^{e^{\overline{a}}\infty} (e^{-a }u)^{-s-1} e^{-u}d(e^{-a}u) =e^{a s}\int_0^{e^{\overline{a}}\infty}+\int_{e^{\overline{a}}\infty}^\infty u^{-s-1} e^{-u}du= e^{a s} \Gamma(-s)$$

For $Re(s) \in (-1,0)$ and $a =b+ i\pi/2$ then $$2i\int_0^\infty t^{-s-1} \sin(t) dt = \lim_{b \to 0^+} \int_0^\infty t^{-s-1} (e^{-e^{b+i\pi/2} t}-e^{-e^{b-i\pi/2} t})dt = \lim_{b \to 0^+}(e^{s(b+i\pi/2) }-e^{s(b-i\pi/2)}) \Gamma(-s)= 2i \sin(\pi s/2)\Gamma(-s)$$

And $\int_0^\infty t^{-s-1} \sin(t) dt =\sin(\pi s/2)\Gamma(-s)$ stays true for $\Re(s) \in(-1,1)$ by analytic continuation

Note a similar derivation with $\int_0^\infty t^{s-1} \log(1-e^{-t})dt$ yields the functional equation for $\zeta(s)$, as $Im(\log(1-e^{4i \pi t})) = 2i\pi t - 2i\pi\lfloor t \rfloor$


I thought it might be instructive to present an approach that uses Laplace Transforms, an integral representation of the Beta Function, the relationship between the Beta Function and Gamma Function, and Euler's Reflection Formula for the Gamma Function. To that end we now proceed.

Let $f(x)=\sin(x)$ and $g(x)=\frac{1}{x^{s+1}}$. Then, the Laplace Transform of $f$ is

$$\mathscr{L}\{f\}(x)=\frac{1}{x^2+1}\tag1$$

and for $\text{Re}(s)\in(-1,0)$, the inverse Laplace Transform of $g$ is

$$\mathscr{L}^{-1}\{g\}(x)=\frac{x^s}{\Gamma(s+1)}\tag2$$

Using $(1)$ and $(2)$ we see that

$$\begin{align} \int_0^\infty \frac{\sin(y)}{y^{s+1}}\,dy&=\frac1{\Gamma(s+1)}\int_0^\infty \frac{x^s}{x^2+1}\,dx\\\\ &=\frac{1}{2\Gamma(s+1)}\int_0^\infty \frac{x^{(s-1)/2}}{1+x}\,dx\\\\ &=\frac1{2\Gamma(s+1)}B\left(\frac{1+s}{2},\frac{1-s}{2}\right)\\\ &=\frac{\Gamma\left(\frac{1+s}{2}\right)\Gamma\left(\frac{1-s}{2}\right)}{2\Gamma(s+1)}\\\\ &=\frac{\frac{\pi}{\cos(\pi s/2)}}{2\frac{\pi}{\Gamma(-s)\sin(\pi(s+1))}}\\\\ &=-\Gamma(-s)\sin(\pi s/2) \end{align}$$

as expected!

See THIS ANSWER for reference.

Tags:

Analysis

Number Theory

Real Analysis

Complex Analysis

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