I am trying to show that $$\int_0^{\pi/2} \frac {\sin(u+a\tan u)} {\sin u}\,\mathrm d u=\frac {\pi} 2$$

Can this be done via contour integration? I'm not really sure which contour to pick. I have tried substitutions like $\pi/2 - u$ but they haven't helped. I have tried differentiating with respect to $a$ too. I got $I'(a)=\int_0^{\pi/2} \frac {\cos(u+a\tan u)} {\cos u}\,\mathrm d u$ And I know $I(a)=\int_0^{\pi/2} \frac {\cos(u-a\cot u)} {\cos u}\,\mathrm d u$

This came up on an undergraduate end of year exam so any solution shouldn't be too advanced.

  • $\begingroup$ What is your $a$? $\endgroup$ – Gibbs Jun 14 '18 at 13:45
  • $\begingroup$ $a$ is any number . I think $\endgroup$ – user544680 Jun 14 '18 at 13:47
  • $\begingroup$ Perhaps $u=\arctan(v)$ is a productive substitution? It might help to have an idea of where this came from, because it's not entirely clear to me that complex variable methods are the right way to go. But of course they are if it is from a complex analysis class. In any case you have a really really bad singularity at $\pi/2$, so anything you do will need to regularize that... $\endgroup$ – Ian Jun 14 '18 at 13:50
  • $\begingroup$ Yes, an equivalent integral is $\int_0^\infty \frac{\sin(arctan(x)+ax)}{x\sqrt{1+x^2}} dx$ which is dramatically less singular. Now a natural thing to try is to show that $I(a)$ is constant and then evaluate $I(0)$ separately. $\endgroup$ – Ian Jun 14 '18 at 13:55
  • $\begingroup$ @Ian maybe not. I'm not sure. This isn't from a complex analysis exam. It is an old tripos question from 1881. $\endgroup$ – user544680 Jun 14 '18 at 13:59

$$I(a)=\int_0^{\pi/2}\frac{\sin(u+a\tan u)}{\sin u}du$$

by use this identitie $$\sin(x+y)=\sin x \cos y + \cos x \sin y$$ so $$I(a)=\int_0^{\pi/2}(\cos(a \tan u)+\frac{\sin(a\tan u)}{\tan u})du$$ let $$u=\tan^{-1}v$$ then $$I(a)=\int_0^{\infty}\frac{\cos(av)+\frac{\sin(av)}{v}}{(1+v^2)}dv$$ differentiate both side with respect to $a$ $$I'(a)=\int_0^{\infty}\frac{\cos(av)-v \sin (av)}{(1+v^2)}dv$$

both of

$$\int_0^{\infty}\frac{\cos(av)}{(1+v^2)}dv ,,,,,,,,,\int_0^{\infty}\frac{v \sin (av)}{(1+v^2)}dv$$

can be shown here and here in real method

thats lead to $$I'(a)=0$$

$$\Rightarrow I(a) \text{ is const}$$


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  • $\begingroup$ This method makes it really obvious that $I(0)=\pi/2$,+1. $\endgroup$ – Ian Jun 15 '18 at 3:36

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