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Given $\sinh{(x)}\sinh{(y)}=1$, I have to find the integral: $$\int_{0}^{+\infty}y\,dx=\dfrac{\pi^2}{4}.$$

I try to use the fact that this $$(e^x-e^{-x})(e^y-e^{-y})=4$$ut i have no idea of how to get $\int ydx=?$

Jack D'Aurizio
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    $y \mapsto \sinh(y)$ is an invertible function. Try to express its inverse in terms of elementary functions, and you will obtain an expression to use for $y$ in the integrand. – G. H. Faust Apr 17 '15 at 07:40
  • Oh，But I think this is hardly –  Apr 17 '15 at 07:41
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    The identity $\sinh(x)\sinh(y)=1$ yields $\sinh(y)dx=dy$ hence the integral is $$\int_0^\infty\frac{ydy}{\sinh(y)}.$$ Now expand $$\frac1{\sinh(y)}=2e^{-y}(1-e^{-2y})^{-1}=2\sum_{n\geqslant0}e^{-(2n+1)y},$$ and integrate term by term, using that $$\int_0^{\infty}ye^{-(2n+1)y}dy=\frac1{(2n+1)^2},$$ to reach $$I=2\sum_{n\geqslant0}\frac1{(2n+1)^2},$$ and conclude. – Did Apr 17 '15 at 07:46
  • @Did,why $\sinh{(y)}dx=dy$? can you explain How did you find it? –  Apr 17 '15 at 07:51
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    Since $\sinh(x)\sinh(y)$ is constant, $\sinh(x)\cosh(y)dy=\cosh(x)\sinh(y)dx$. Now, simplify, using $\sinh(x)=1/\sinh(y)$ and $\cosh(x)=\cosh(y)/\sinh(y)$. – Did Apr 17 '15 at 07:54
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    @Did, you should put that as answer. You make the hard looking integral so simple! – achille hui Apr 17 '15 at 07:56
  • @achillehui Thanks. Even better would be that the OP (or anybody else) writes down a full solution, expanding on the sketch in my comment. – Did Apr 17 '15 at 07:58
  • @Did, I agree that will be even better. Let's wait and see. – achille hui Apr 17 '15 at 08:00

1 Answers1

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We have to compute: $$ I = \int_{0}^{+\infty}\operatorname{arcsinh}\left(\frac{1}{\sinh x}\right)\,dx = \int_{0}^{+\infty}\frac{\operatorname{arcsinh}\frac{1}{u}}{\sqrt{1+u^2}}\,du$$ that is (by replacing $u$ with $\tan t$): $$ I = \int_{0}^{+\infty}\frac{\log(\frac{1+\sqrt{1+u^2}}{u})}{\sqrt{1+u^2}}\,du =\int_{0}^{\pi/2}\frac{\log\left(\frac{1+\cos t}{\sin t}\right)}{\cos t}\,dt$$ or (by using Weierstrass substitution): $$ I = 2\int_{0}^{\pi/4}\frac{\tan^2 w+1}{\tan^2 w-1}\,\log\tan w\,dw=2\int_{0}^{1}\frac{\log v}{v^2-1}\,dv$$ and finally: $$\int_{0}^{1}\frac{-\log v}{1-v^2}\,dv = \sum_{k\geq 0}\int_{0}^{1}(-\log v)v^{2k}\,dv = \sum_{k\geq 0}\frac{1}{(2k+1)^2} = \frac{3}{4}\zeta(2)=\frac{\pi^2}{8}.$$ To summarize the change of variables in just one step (as suggested by Did), we used: $$ v=\tan\left(\frac{\arctan\sinh x}{2}\right).$$

Jack D'Aurizio
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  • @Did: In the second formula, I replaced $u$ with $\tan t$. Going from the second to the third line, I used Weierstrass substitution. Everything's pretty kosher, imho. – Jack D'Aurizio Apr 17 '15 at 09:54
  • Better now. If only by curiosity, you could try to summarize the five changes of variable your solution uses, namely, $x\to u\to t\to x\to v\to-\log v$, into a single one. The result is enlightening. – Did Apr 17 '15 at 10:02
  • @Did: all right. All in once, $$ v=\tan\left(\frac{\arctan\sinh x}{2}\right).$$ – Jack D'Aurizio Apr 17 '15 at 10:04
  • Maybe, but who cares? Note rather that $w=-\log v$ solves $$\sinh(x)\sinh(w)=1,$$ and compare with the direct method in my comment. – Did Apr 17 '15 at 10:06
  • @Did: I saw your solution in the comments, but what is the point in underlying it here? I just took a slightly different path. – Jack D'Aurizio Apr 17 '15 at 10:08
  • The point is that it took you five, seemingly random, steps to reach the exact change of variable the condition $\sinh(x)\sinh(y)=1$ imposes from the start. And the problem one can have with this (and that I honestly thought you would have had yourself once you realized the thing) is that this détour makes nearly impossible to see what really happens here (and why the whole shebang works, in the end). – Did Apr 17 '15 at 10:13
  • @Did: Nothing's random. I wanted to reach an integral resembling a derivative of a Beta function, and I did it by following the most natural (to me) manipulations. Clarity is in the eye of the beholder. – Jack D'Aurizio Apr 17 '15 at 10:16
  • @Did: and by the way, why don't you just post your solution as an answer, instead of criticizing mine? – Jack D'Aurizio Apr 17 '15 at 10:17
  • "I did it by following the most natural (to me) manipulations." Sure, and there is absolutely nothing wrong with that (who said there was, I wonder?). But once the computations are over, once the dust has settled, so to speak, why not look back at the path used and think about what happened and, once again, about what made the whole thing work? It seems obvious to me that this can only bring more understanding -- but if you think this brings no more "clarity" to note that your $x\to u\to t\to x\to v\to-\log v$ actually amounts to $x\to y$, no problem, let us agree to disagree. – Did Apr 17 '15 at 10:26