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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.

  • What is your $a$? – Gibbs Jun 14 '18 at 13:45
  • $a$ is any number . I think –  Jun 14 '18 at 13:47
  • 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... – Ian Jun 14 '18 at 13:50
  • 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. – Ian Jun 14 '18 at 13:55
  • @Ian maybe not. I'm not sure. This isn't from a complex analysis exam. It is an old tripos question from 1881. –  Jun 14 '18 at 13:59
  • Yes I think my suggestion is the way to go, you can get the actual antiderivative of $\frac{\sin(\arctan(x))}{x\sqrt{1+x^2}}$, integrate that to get $\pi/2$ easily, and now you should show that $I'(a)=0$. – Ian Jun 14 '18 at 14:01
  • Ah Thanks. I'll try this –  Jun 14 '18 at 14:03
  • @Ian: $$\frac{d}{da} \int_0^\infty \frac{\sin (\arctan x + ax)}{x \sqrt{1 + x^2}}, dx = \int_0^\infty \frac{\cos (\arctan x + ax)}{\sqrt{1+x^2}}, dx$$ and I don't see an easy way of proving that that this is zero. – Connor Harris Jun 14 '18 at 14:06
  • @Connor if you expand out the $\cos$ you get something that I believe is solvable by contour integration –  Jun 14 '18 at 14:10
  • @user1488 The integrand has even symmetry, in case this is helpful. – Connor Harris Jun 14 '18 at 14:15
  • Done it. You end up with $\int_0^{\infty} \frac{\cos{av}}{1+v^2} ,\mathrm d v - \int_0^{\infty} \frac{v\sin{av}}{1+v^2} ,\mathrm d v$. Each integral evaluates to $\pi e^{-a}/2 $ by integrating over a semi circle in the upper half plane. Thanks for the help –  Jun 14 '18 at 14:23
  • @user1488 post this as a full answer, please! – Connor Harris Jun 14 '18 at 14:27
  • Yes, please do write a full answer. It's totally within "protocol" to answer your own question here. Not so much to ask a question that you already know the answer to and then immediately answer it (use a blog or something for that), but if the comments give you the hint you need, we appreciate self-answered questions. – Ian Jun 14 '18 at 15:46
  • @Gibbs usually if a variable ($a$ for example) is not described, it is assumed to be a real number, id est $a\in\mathbb{R}$. – Mr Pie Jun 15 '18 at 01:09
  • @user477343 when a variable or a parameter is not declared, asking where it sits is a natural question. I do not take anything for granted. – Gibbs Jun 15 '18 at 08:21

1 Answers1

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

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 $$I'(a)=0$$

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

$$I(a)=I(0)=\pi/2$$

mnsh
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