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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:

\begin{equation} \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) \end{equation}

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 \begin{equation} 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!} \end{equation}

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: \begin{equation} \Gamma(s)\Gamma(1 - s) = \frac{\pi}{\sin(\pi s)} \rightarrow \Gamma(1 - s) = \frac{\pi}{\Gamma(s)\sin(\pi s)} \end{equation}

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.

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    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 '19 at 00:46
  • @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. –  Feb 14 '19 at 00:48
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    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 '19 at 00:54
  • @reuns - I was hoping to avoid complex analysis in all forms (if possible). Do you know if that is possible? –  Feb 14 '19 at 01:08
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    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 '19 at 03:34

3 Answers3

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

reuns
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As pointed out in the comments, here has indeed a nice perspective to solve this, but sadly due to Euler Reflection Formula it isn't really the "Real Analysis" way.

To explain this, the integral $\int_0^{\infty}x^{s-1}\sin (x)dx$ converges iff $s \in (-1,1)$, else it diverges.

But, this can be viewed as the Mellin Transform of $\sin (x)$ that can be analytically extended for any complex $s$ barring a few set of values (which are the non-negative integers here) via the RMT. So, the RMT results a meromorphic function, and hence is a tool of Complex Analysis.

Even the Euler Reflection Formula is a result of complex analysis, since you're working with the meromorphic $\Gamma (z)$ as we use its Weierstrass product in the standard derivation. ProofWiki uses the Weierstrass products of $\Gamma (z)$ and $\sin (z)$ respectively to derive the formula.

Even if one restricts $s \in (0,1)$ and tries to derive the Euler Reflection Formula via $B(s,1-s) = \Gamma(s) \Gamma(1-s)$, the resulting integral from the definition of Beta Function invariably invokes the use of Weierstrass products in some way or the other to get a closed form, or use Fourier series (which also uses Complex Analysis to determine the coefficients) to get a certain series representation of $\pi \csc (\pi x)$.

Idris Lokhande
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B the formula $$ \int_0^{\infty} x^{s-1} e^{-t x} d x=\frac{\Gamma(s)}{t^s}, $$ we have $$ \begin{aligned} \mathcal{M}(\sin x)(s) & =\int_0^{\infty} x^{s-1} \sin x d x \\ & =-\Im \int_0^{\infty} x^{s-1} e^{-ix } d x \\ & =-\Im \left( \frac{\Gamma(s)}{i^s}\right) \\ & =-\Gamma(s) \Im\left(e^{-\frac{\pi}{2} s i}\right) \\ & =-\Gamma(s) \Im\left[\cos \left(\frac{\pi}{2} s\right)-i \sin \left(\frac{\pi}{2} s\right)\right] \\ & =\Gamma(s) \sin \left(\frac{\pi}{2} s\right) \end{aligned} $$

Lai
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