Showing that integral is related to sine function in elementary means 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 :) 
 A: 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.

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

A: 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.
