# Proving $\mathcal{M}\left(\sin(x)\right)(s) = \Gamma(s)\sin\left(\frac{\pi}{2}s \right)$ using Real Analysis

recently I've been investigating Mellin Transforms and this morning solved for case of $$\sin(x)$$ using Ramunajan's Master Theorem. I was curious if there were any Real based methods to evaluate this integral as in searching around online all proofs seem to rely on Contour Integration.

My approach:

$$$$\mathcal{M}\left(\sin(x)\right)(s) = \int_0^\infty x^{s - 1}\sin(x)\:dx = \Gamma(s)\sin\left(\frac{\pi}{2}s \right)$$$$

By definiton:

\begin{align} \mathcal{M}\left(\sin(x)\right)(s) &= \int_0^\infty x^{s - 1}\sin(x)\:dx = \int_0^\infty x^{s - 1}\left[\sum_{m = 0}^{\infty} (-1)^m \frac{x^{2m + 1}}{(2m + 1)!} \right]\:dx \\&= \int_0^\infty x^{s} \sum_{m = 0}^{\infty} (-1)^m \frac{\left(x^2\right)^m}{(2m + 1)!}\:dx \end{align}

Here we make the substitution $$u = x^2$$: \begin{align} \mathcal{M}\left(\sin(x)\right)(s) &= \int_0^\infty \left(\sqrt{u}\right)^{s} \sum_{m = 0}^{\infty} (-1)^m \frac{\left(u\right)^m}{(2m + 1)!}\frac{du}{2\sqrt{u}} = \frac{1}{2}\int_0^\infty u^{\frac{s - 1}{2}}\sum_{m = 0}^{\infty} (-1)^m \frac{u^m}{(2m + 1)!}\:du \\ &= \frac{1}{2}\int_0^\infty u^{\frac{s + 1}{2} - 1}\sum_{m = 0}^{\infty} \frac{(-u)^m}{(2m + 1)!}\:du = \frac{1}{2}\mathcal{M}\left(g(u)\right)\left(\frac{s + 1}{2}\right) \end{align}

Where $$$$g(u) = \sum_{m = 0}^{\infty} \frac{(-u)^m}{(2m + 1)!} = \sum_{m = 0}^{\infty} \frac{m!}{(2m + 1)!}\frac{(-u)^m}{m!} = \sum_{m = 0}^{\infty} \frac{\Gamma(m + 1)}{\Gamma(2m + 2)}\frac{(-u)^m}{m!}$$$$

We observe that this form enables the use of Ramanujan's Master Theorem. Thus: \begin{align} \mathcal{M}\left(\sin(x)\right)(s) &= \frac{1}{2}\mathcal{M}\left(g(u)\right)\left(\frac{s + 1}{2}\right) = \frac{1}{2} \Gamma\left( \frac{s + 1}{2}\right) \cdot \frac{\Gamma\left(- \frac{s + 1}{2} + 1\right)}{\Gamma\left(2\cdot -\frac{s + 1}{2} + 2\right)} \\ &= \frac{1}{2} \frac{\Gamma\left( \frac{s + 1}{2}\right)\Gamma\left(1 - \frac{s + 1}{2}\right)}{\Gamma(1 - s)} \end{align}

Employing Euler's Reflection Formula on the terms in the numerator we arrive at: \begin{align} \mathcal{M}\left(\sin(x)\right)(s) &= \frac{1}{2} \frac{\Gamma\left( \frac{s + 1}{2}\right)\Gamma\left(1 - \frac{s + 1}{2}\right)}{\Gamma(1 - s)} = \frac{1}{2} \cdot \frac{1}{\Gamma(1 - s)}\cdot \frac{\pi}{\sin\left(\pi \cdot \frac{s + 1}{2}\right)} = \frac{\pi}{2} \frac{1}{\Gamma\left(1 - s\right)\sin\left(\frac{\pi}{2} (s + 1)\right)} \end{align}

Now Euler's Reflection Formula: $$$$\Gamma(s)\Gamma(1 - s) = \frac{\pi}{\sin(\pi s)} \rightarrow \Gamma(1 - s) = \frac{\pi}{\Gamma(s)\sin(\pi s)}$$$$

Returning to our integral and observing that $$\sin\left(\frac{\pi}{2} (s + 1)\right)= \cos\left(\frac{\pi}{2}s\right)$$ we arrive at:

\begin{align} \mathcal{M}\left(\sin(x)\right)(s) &= \frac{\pi}{2} \frac{1}{\Gamma\left(1 - s\right)\sin\left(\frac{\pi}{2} (s + 1)\right)} = \frac{\pi}{2} \cdot \frac{1}{\frac{\pi}{\Gamma(s)\sin(\pi s)} \cdot \cos\left(\frac{\pi}{2}s\right)} = \frac{\Gamma(s)\sin(\pi s)}{2 \cos\left(\frac{\pi}{2} s\right)} \end{align}

We now use the double angle formula $$\sin\left(\pi s\right) = 2\sin\left(\frac{\pi}{2} s\right)\cos\left(\frac{\pi}{2} s\right)$$ to yield:

\begin{align} \mathcal{M}\left(\sin(x)\right)(s) &= \frac{\Gamma(s)\sin\left(\pi s\right)}{2 \cos\left(\frac{\pi}{2} s\right)} = \frac{\Gamma(s)\cdot 2\sin\left(\frac{\pi}{2} s\right)\cos\left(\frac{\pi}{2} s\right)}{2 \cos\left(\frac{\pi}{2} s\right)} = \Gamma(s)\sin\left(\frac{\pi}{2} s\right) \end{align}

As required.

• Ramanujan's Master Theorem + Euler's Reflection Formula + ... is the exact opposite of "using real analysis". The only way to show your claim is to start with $\int_0^\infty x^{s-1} e^{-ax}dx = a^{-s} \Gamma(s)$ for $\Re(s) > 0$, from there $\int_0^\infty x^{s-1} \sin(x)dx = \lim_{b \to 0^+} \int_0^\infty x^{s-1} \frac{e^{-(b+i)x}-e^{-(b-i)x}}{2i}dx = \lim_{b \to 0^+} \frac{(b+i)^{-s}-(b-i)^{-s}}{2i} \Gamma(s)= \sin(\pi s/2) \Gamma(s)$ for $\Re(s) \in (0,1)$ – reuns Feb 14 at 0:46
• @reuns - yes indeed, I realised my wording was incorrect and have corrected. I'm curious whether this integral can be solved using real based methods. – user150203 Feb 14 at 0:48
• To avoid complex integrals you can say for $a > 0$, $\int_0^\infty x^{s-1} e^{-ax}dx - a^{-s} \Gamma(s)=0$. Since the LHS is analytic in $a, \Re(a) > 0$ (binomial series + expand $e^{-(a+b)x}=e^{-ax}e^{-bx}$ in $a$) the equality stays true for every $a, \Re(a) > 0$. – reuns Feb 14 at 0:54
• @reuns - I was hoping to avoid complex analysis in all forms (if possible). Do you know if that is possible? – user150203 Feb 14 at 1:08
• Here is an essentially real method I gave to a very old question of finding the Mellin transform for $\sin x$ only a week ago! It uses properties of the Laplace transform. I say "essentially" a real method since while there may be no appearance of the imaginary unit $i$, it still relies on Euler's reflexion formula for the Gamma function. – omegadot Feb 14 at 3:34

## 1 Answer

For $$b > 0, \Re(s) > 0$$ $$b^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-bx}dx$$

By induction if for some $$b,\Re(b) > 0$$ we have $$\forall s, \Re(s) >0,b^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-bx}dx$$ then for every $$|a/b|< 1, \Re(a+b) > 0$$, using the binomial series $$\int_0^\infty x^{s-1} e^{-(a+b)x}dx = \sum_{k=0}^\infty \int_0^\infty x^{s-1}\frac{(-ax)^k}{k!} e^{-bx}dx=\sum_{k=0}^\infty \frac{(-a)^k}{k!} b^{-s-k} \Gamma(s+k)\\ =b^{-s} \Gamma(s)\sum_{k=0}^\infty {-s \choose k} (a/b)^k = (a+b)^{-s} \Gamma(s)$$

From which $$b^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-bx}dx$$ is true for every $$b, \Re(b) > 0$$

And hence for $$\Re(s) \in (0,1)$$ $$\int_0^\infty x^{s-1} \sin(x)dx = \lim_{b \to 0^+} \int_0^\infty x^{s-1} \frac{e^{-(b-i)x}-e^{-(b+i)x}}{2i}dx \\ = \lim_{b \to 0^+} \frac{(b-i)^{-s}-(b+i)^{-s}}{2i} \Gamma(s)= \sin(\pi s/2) \Gamma(s)$$

• I don't want to be critical, but this does use complex analysis. The complex defintion of sin(x) requires complex analysis. – Highvoltagemath Aug 14 at 23:32