# Old & cool integral $\int_0^{\pi} \sin^{b-1}(x) \sin(a x) \ dx=\frac{\pi \sin(a \pi/2)}{2^{b-1}b B\left(\frac{b+a+1}{2},\frac{b-a+1}{2}\right)}$

Here is an integral that appears in the table of integrals by Gradshtein and Ryzhik, it was also studied by Ramanujan (not sure his original solution was found - it seems it doesn't appear in
any of the notebooks).

$$\int_0^{\pi} \sin^{b-1}(x) \sin(a x) \ dx=\frac{\pi \sin(a \pi/2)}{2^{b-1}b B\left(\frac{b+a+1}{2},\frac{b-a+1}{2}\right)}$$

Now, by complex analysis, one can brifely finish it, I'm not interested in such a solution. But thinking of Ramanujan I'm sure he had a solution using methods of real analysis (and to avoid
possible misunderstandings, I mean not even a touch on complex numbers - to be clear).

Do you know such a solution? Post it only if you want to, I'm only curious if such solutions are known, maybe some simple such solutions?

Application of the integral above (supplementary question)

Prove that

$$\int_0^{\pi/2} \frac{\log (\sin (x))+x \csc ^2(x)-x \cot (x)}{x^2+\log ^2(\sin (x))} \, dx=\text{Si}\left(\frac{\pi }{2}\right),$$

http://mathworld.wolfram.com/SineIntegral.html

or simply show that

$$\int_0^{\pi/2} \frac{x \cot (x)-\log (\sin (x))}{x^2+\log ^2(\sin (x))} \, dx=\frac{\pi}{2}.$$

• Have you tried using the Fourier series expansion of $\sin(ax)$ ? – TheOscillator Mar 26 '16 at 20:37
• If it appears in G&R, it'd help to reference where in the text it shows up. Additionally, for what $a,b$ does this identity hold? – Semiclassical Mar 27 '16 at 20:31
• The properties of Beta and Gamma functions are closely related to complex analysis, and can't be easily obtained another way. So can we use those without proof? – Yuriy S Aug 18 '16 at 8:02