Evaluate$$\int_0^1 \dfrac{\ln \left(a+\sqrt{a^2+1}\right)}{a\sqrt{a^2+1}}da$$

Unfortunately I have absolutely no idea as to how to approach this problem. It was suggested that I try Integration By Parts, but I couldn't even understand what to do with it. I'm really sorry for not being able to give any more inputs of my own, but I just don't know what to do. $$$$I would be truly grateful if somebody could spare the time to show me how to evaluate this Integral. Many, many thanks in advance!

Note: Please could you not use Hyperbolic Functions? Unfortunately I haven't learnt to use them yet.

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    $\begingroup$ Can you show us where you found this problem ? Because, the answer does not seem to have a closed form without the use of specials functions (for example polylogarithms), so it is not some basic calculus. I've read your comment where you said you didn't see hyperbolic functions yet, and the calculation without it must not be simple, so I'm surprised that you're asked something like that. $\endgroup$ – Sylvain L. Jun 23 '15 at 20:21
  • $\begingroup$ Sir, I was trying to solve $$\int_0^{\pi/2}\dfrac{x \cos(x)}{1+\sin^2(x)} dx$$ Using Differentiation Under the Integral Sign, I reduced it to $\dfrac{\pi^2}{8}-J(1)$ where $$J(a)=\int_0^1 \dfrac{\ln \left(a+\sqrt{a^2+1}\right)}{a\sqrt{a^2+1}}da$$ $\endgroup$ – Ishan Jun 23 '15 at 20:25
  • $\begingroup$ @SylvainL. Sir, the closed form is given as $$ \dfrac { { log }^{ 2 }(1+\sqrt { 2 } ) }{ 2 }$$ $\endgroup$ – Ishan Jun 23 '15 at 20:27
  • $\begingroup$ Yeah, it seems right numerically. I guess there are some relations between the polylogarithm that gives this. $\endgroup$ – Sylvain L. Jun 23 '15 at 20:32
  • $\begingroup$ Exactly, it follows from combining my answer with identities $(4),(5),(7)$ here. $\endgroup$ – Jack D'Aurizio Jun 23 '15 at 20:34

If we use the substitution $a=\sinh u$, then $u=\log v$, we are left with:

$$ I = \int_{0}^{\log(1+\sqrt{2})}\frac{u\,du}{\sinh u}=\int_{1}^{1+\sqrt{2}}\frac{2\log v}{v^2-1}\,dv = \left.\frac{d}{d\alpha}\int_{1}^{1+\sqrt{2}}\frac{v^{\alpha}\,dv}{v^2-1}\right|_{\alpha=0}$$ that ultimately depends on $\frac{\pi^2}{4}$, the product of some logarithms, $\text{Li}_2(1-\sqrt{2})$ and $\text{Li}_2(\sqrt{2}-1)$:

$$ I = \frac{\pi^2}{4}+\log(\sqrt{2}+1)\log(\sqrt{2}-1)+\text{Li}_2(1-\sqrt{2})-\text{Li}_2(\sqrt{2}-1),$$

since: $$ \int\frac{\log v}{v+1}\,dv=\log v \log(1+v)+\text{Li}_2(-v), $$ $$ \int\frac{\log v}{v-1}\,dv = -\text{Li}_2(1-v). $$ Not exactly a trivial integral. The closed form:

$$ I = \frac{\log^2(1+\sqrt{2})}{2} $$

follows for the functional identity for the dilogarithm: $$ \text{Li}_2(1-x)+\text{Li}_2(1-x^{-1})=-\frac{1}{2}\log^2 x $$

that is straightforward to prove through differentiation.

  • $\begingroup$ Are you sure about the $1+v^2$ in the denominator? Shouldn't that be $v^2-1$? $\endgroup$ – mickep Jun 23 '15 at 20:21
  • $\begingroup$ Yet another place :) But +1 for this anyways... (The first two substitutions could be combined as one, $a=(v^2-1)/(2v)$, that is $v=a+\sqrt{1+a^2}$.) $\endgroup$ – mickep Jun 23 '15 at 20:22
  • $\begingroup$ @mickep: true again. Now everything should be right. $\endgroup$ – Jack D'Aurizio Jun 23 '15 at 20:24
  • $\begingroup$ @JackD'Aurizio Thanks Sir. But unfortunately I don't know Hyperbolic Functions, Sir. Also Sir, the Closed form of the original Integral, $$\int_0^{\pi/2} \dfrac{x\cos(x)}{1+\sin^2(x)} dx$$ comes out to be $$ \dfrac { { log }^{ 2 }(1+\sqrt { 2 } ) }{ 2 }$$ $\endgroup$ – Ishan Jun 23 '15 at 20:32
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    $\begingroup$ @JackD'Aurizio : Quite nice answer, I upvote. Still a little bit surprised that no-one found a way to get the suare log without going through polylogs, but anyway, your method is efficient. $\endgroup$ – Sylvain L. Jun 23 '15 at 21:05

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