# Integral with parameter

Is it possible to express in a closed form the integral $$\int_{0}^{\pi/2}\frac{\sin \left ( ax \right )}{\sin x+\cos x}\, {\rm d}x,\,\,\, a\in \mathbb{N}$$

Well, I find it very difficult. Well, I know how to express the integral $$\int_{0}^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,{\rm d}x=\int_{0}^{\pi/2}\frac{\cos x}{\sin x+\cos x}\, {\rm d}x=\frac{\pi}{4}$$ by applying the sub $u=\frac{\pi}{2}-x$ , but in general I don't have a clue.

If someone could help me , that would be nice!

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Nice question. Wolfram Alpha can't find a closed formula. –  Siminore Jul 20 '14 at 10:58
After a substitution $t=x-\pi/4$ we get $I=\sqrt2\sin\frac{\pi a}{4}\int_0^{\pi/4}\frac{\cos ax}{\cos x}\,dx$. The last integral could be treated using recurrence relation $\cos ax=2\cos x\cos(a-1)x-\cos(a-2)x$. I hope this might help. –  CuriousGuest Jul 20 '14 at 11:14
@CuriousGuest Well, how possible is to evaluate the last integral? I think I've seen it somewhere before... Nice approach though, thnx! –  Tolaso Jul 20 '14 at 11:25
@Tolaso It's easy to integrate it from 0 to $\pi$ (in this case $I_a=I_{a-2}$), but for segment $[0,\pi/4]$ the recurrence relation isn't so nice. –  CuriousGuest Jul 20 '14 at 11:28
@CuriousGuest That is why I asked that. Therefore , I don't know.. I only know the integrals you've mentioned $\int_0^\pi \frac{sin ax}{\sin x}\, dx$ and that other with cosine. Anyway, thanks a lot for your answer. I'll take a look at it. –  Tolaso Jul 20 '14 at 11:35

If you are really interested in a closed form formula then let's consider the integral

$$I = \int_{0}^{\pi/2} \frac{e^{inx}}{\cos(x)+\sin(x)}dx$$

where your integral equals to the imaginary part of $I$. $I$ can have the closed form in terms of the Lerch zeta function

$$I = \frac{(1-i) e^{\frac{i\pi n}{2}}} {2}\left( \Phi \left( -i,1,\frac{n+1}{2}\right) - \Phi \left( i,1, \frac{n+1}{2} \right ) \right) .$$

Note:

1) Maple $17$ can not give an answer for this integral! I do not know about Maple 18. Already a different form for the answer, computed by Mathematica $9$, was posted.

2) The real part of $I$ evaluates the integral

$$\int_{0}^{\pi/2}\frac{\cos(nx)}{\cos(x)+\sin(x)}dx.$$

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This one is beautiful ! Thanks for this answer –  Claude Leibovici Jul 20 '14 at 13:11
I have a problem : for $n=1$ the result from your formula seems to be $\frac{\log (2)}{2}$ –  Claude Leibovici Jul 20 '14 at 13:19
@ClaudeLeibovici: Thanks for the comment. I really appreciate it. –  Mhenni Benghorbal Jul 20 '14 at 13:19
@ClaudeLeibovici: It gave me $\pi/4+i\pi/4$. Make sure you copied it correctly. –  Mhenni Benghorbal Jul 20 '14 at 13:21
@MhenniBenghorbal Great ! Thank you all for your answers. –  Tolaso Jul 20 '14 at 13:31

You have an answer from Vladimir but I had fun computing the integral for some values of $n$ and I let that for your curiosity.

Let $$I_n=\int_{0}^{\pi/2}\frac{\sin \left ( ax \right )}{\sin x+\cos x}\, {\rm d}x$$ $$I_1=\frac{\pi }{4}$$ $$I_2=2-\sqrt{2} \tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right)$$ $$I_3=1-\frac{\pi }{4}$$ $$I_4=0$$ $$I_5=1-\frac{\pi }{4}$$ $$I_6=\sqrt{2} \tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{14}{15}$$ $$I_7=\frac{\pi }{4}-\frac{2}{3}$$ $$I_8=0$$ $$I_9=\frac{\pi }{4}-\frac{2}{3}$$ $$I_{10}=\frac{454}{315}-\sqrt{2} \tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right)$$ $$I_{11}=\frac{13}{15}-\frac{\pi }{4}$$ $$I_{12}=0$$ We can understand that the general formula must be far away from simple.

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Well, I had computed the three first values of the integral (that is for $a=1, 2, 3$) got the same results.. and then I gave up. OK, Vladimirs answer is good but I think that a closed form may not exist... What I observe now , from your post, is that every 4 four values the integral equals zero. That may gives us a pattern.. who knows? –  Tolaso Jul 20 '14 at 12:58
There is closed form ... which is a nightmare (at least, to me). The fact that every four values the integral equals zero is obvious since the numerator is $\sin(4kx)$ and the integration is between $0$ and $\frac{\pi }{2}$. Good luck and enjoy ! Cheers :) –  Claude Leibovici Jul 20 '14 at 13:02
I'd imagine that it'd simplify things somewhat to divide into cases of $n$ mod 4 owing to the distinct difference in behaviors. –  Semiclassical Jul 20 '14 at 14:20

This is $$-\Biggl(\left(\frac{1}{2}+\frac{i}{2}\right) e^{-\frac{1}{2} i \pi n} \biggl(e^{\frac{i \pi n}{2}} \biggl((n-1) \biggl(e^{\frac{i \pi n}{2}} \, _2F_1\biggl(1,\frac{n+1}{2};\frac{n+3}{2};-i\biggr)\\+i \, _2F_1\biggl(1,\frac{n+1}{2};\frac{n+3}{2};i\biggr)\biggr)+i (n+1) \, _2F_1\biggl(1,\frac{1-n}{2};\frac{3-n}{2};i\biggr)\biggr)\\+(n+1) \, _2F_1\biggl(1,\frac{1-n}{2};\frac{3-n}{2};-i\biggr)\biggr)\Biggr)/(n^2-1)$$ according to Wolfram Mathematica$^{TM}$ 9

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