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Problem:

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

It was suggested that I try Integration By Parts.

User1234
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    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. – Sylvain Lacroix Jun 23 '15 at 20:21
  • Yeah, it seems right numerically. I guess there are some relations between the polylogarithm that gives this. – Sylvain Lacroix Jun 23 '15 at 20:32
  • Exactly, it follows from combining my answer with identities $(4),(5),(7)$ here. – Jack D'Aurizio Jun 23 '15 at 20:34

2 Answers2

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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.

Jack D'Aurizio
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    Are you sure about the $1+v^2$ in the denominator? Shouldn't that be $v^2-1$? – mickep Jun 23 '15 at 20:21
  • Yet another place :) But +1 for this anyways... (The first two substitutions could be combined as one, $a=(v^2-1)/(2v)$, that is $v=a+\sqrt{1+a^2}$.) – mickep Jun 23 '15 at 20:22
  • @mickep: true again. Now everything should be right. – Jack D'Aurizio Jun 23 '15 at 20:24
  • The original integral is best suited for an evaluation through Integration by parts + Parseval's theorem. The reason for which that combination of dilogarithms gives a squared logarithm are identities $(4),(5),(7)$ here. – Jack D'Aurizio Jun 23 '15 at 20:36
  • Sir, I do not know how to do what you want to do, without using the things you don't know. Probably you should fill the gaps in you mathematical knowledge before trying to solve such hard problems. Otherwise, you will always have to attack them with one hand tied behind your back. – Jack D'Aurizio Jun 23 '15 at 20:47
  • Anyway, have a look at this question – Jack D'Aurizio Jun 23 '15 at 20:54
  • @BetterWorld: and this one, too. – Jack D'Aurizio Jun 23 '15 at 20:58
  • @BetterWorld : I don't know exactly where you find your exercices, if it is for a course or just by yourself in textbooks. If it's the latter, you should learn a few things about hyperbolic trigonometry : it is not very hard (especially if you're used to usual trigo) and it is quite often necessary or at least very useful to tackle some integrals. – Sylvain Lacroix Jun 23 '15 at 21:03
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    @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. – Sylvain Lacroix Jun 23 '15 at 21:05
  • @BetterWorld: I feel I owe you some apology since I acted like you was much older. Given your age there is nothing surprising in you mathematical knowledge to have some gaps, especially if you are a self-learner. I just hope to have given you something to think about, sorry again if I sounded a bit rude. – Jack D'Aurizio Jun 23 '15 at 21:23
  • @BetterWorld: what about starting with this Wikipedia page? – Jack D'Aurizio Jun 24 '15 at 12:25
  • I'm afraid most references I might know are in french, and to be honest, I'm far from being a specialist on hyperbolic trigo ; I mostly remember what I've seen in my different calculus classes but I've never studied it in books. As far as I know, there is not a big theoretical and abstract background for hyp. trigo, it's mostly getting used to the identities, learn how to recognize the derivatives, doing examples of integrals involving the functions and so on. Maybe you can start with references on the internet and see if you can use them to understand the resolution of problem like this post. – Sylvain Lacroix Jun 24 '15 at 12:28
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Rewrite $\ln (a+\sqrt{a^2+1})= \sinh^{-1} a $ \begin{align} I=& \int_0^1 \frac{\ln (a+\sqrt{a^2+1})}{a\sqrt{a^2+1}}da\\ =&\int_{0}^{\infty}\frac{\sinh^{-1} a}{a\sqrt{a^2+1}} \overset{a=\sinh t}{da} -\int_{1}^{\infty}\frac{\sinh^{-1} a}{a\sqrt{a^2+1}} \overset{a\to1/a}{da}\\ =&\int_{0}^{\infty} \frac t{\sinh t}\overset{t=\ln y}{dt}-\int_{0}^{1}\frac{\text{csch}^{-1} a}{\sqrt{a^2+1}} \overset{ibp}{da} \\ =&\ 2\int_0^1 \frac{\ln y}{y^2-1}dy -\text{sinh}^{-1}1\ \text{csch}^{-1}1-I\\ = &\ \frac{\pi^2}8-\frac12 (\sinh^{-1}1)^2 \end{align}

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