Showing that integral is related to sine function in elementary means

Question

So I am trying to prove the reflection formula for the gamma function by showing that $$\int_{0}^{\infty} \frac{v^{s-1}}{1+v}dv=\frac{\pi}{\sin(\pi s)}$$ for $0 < \Re(s) < 1$ , as these two statements are (almost) equivalent. I want to do this with elementary means if possible (I was hoping that it was possible to prove it without actually using complex integration, since the integrand is real, treating s "as if" it was simply real.)

My first attempt was this: assume that

$$\frac{d}{dv}\left \{ \frac{f(v)}{g(v)} \right \}= \frac{v^{s-1}}{1+v}$$ so that $$\frac{f'g-g'f}{g^2}=\frac{v^{s-1}}{1+v}$$ Thus, we have $g(v)=\sqrt{1+v}$ . Multiplying with the denominator yields: $$f'g-g'f=v^{s-1}$$ Or equivalently: $$\sqrt{1+v} f'(v)-\frac{f(v)}{2\sqrt{1+v}}=v^{s-1}$$ I thought about trying to solve this using Laplace transform, but got nowhere. The reason is that I don't know the Laplace transform of $v^{s-1}\sqrt{1+v}$
I also tried expressing $$\frac{v^{s-1}}{1+v}$$ as a Laurent series and using integration term by term, without success. Does anyone know how to prove the given identity (in a way as simple as possible) ?

Thanks, R :)

Answer 1

Your given integral is closely related to the Mellin transform and can be evaluated by using Ramanujan's Master Theorem.

Ramanujan's Master Theorem

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

In order to get there we can expand the fraction as a geometric series

\begin{align*} \int_0^\infty \frac{v^{s-1}}{1+v}\mathrm dv&=\int_0^\infty v^{s-1}\sum_{k=0}^\infty (-v)^k\mathrm dv\\ &=\int_0^\infty v^{s-1}\sum_{k=0}^\infty \frac{\Gamma(k+1)}{k!}(-v)^k\mathrm dv \end{align*}

Now we may use the aforementioned theorem with $s=s$ and $\phi(k)=\Gamma(k+1)$ to obtain

\begin{align*} \int_0^\infty v^{\nu-1}\sum_{k=0}^\infty \frac{\Gamma(k+1)}{k!}(-v)^k\mathrm dv&=\Gamma(s)\Gamma(1-s)\\ &=\frac\pi{\sin(\pi s)} \end{align*}

where we used Euler's Reflection Formula in order to perform the last step.

$$\therefore~\int_0^\infty \frac{v^{s-1}}{1+v}\mathrm dv~=~\frac\pi{\sin(\pi s)}$$

Answer 2

Another method.

Recall the definition of the Beta function: $$\mathrm{B}(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\mathrm{B}(b,a)$$ Then recall the Gamma reflection formula: $$\Gamma(s)\Gamma(1-s)=\frac\pi{\sin\pi s}$$ So with $a=s$ and $b=1-s$, we have $$\int_0^1t^{s-1}(1-t)^{-s}\mathrm dt=\int_0^1t^{-s}(1-t)^{s-1}\mathrm dt=\frac\pi{\sin\pi s}$$ Then use the substitution $x=\frac{1-t}{t}$ to see that $$\int_0^\infty \frac{x^{s-1}}{1+x}\mathrm dx=\int_0^1t^{-s}(1-t)^{s-1}\mathrm dt=\frac\pi{\sin\pi s}$$ As desired.