I want to compute the following integral: $$\raise 1ex{\Large\int} \frac{\sqrt{\ln(x+\sqrt{1+x^2}})}{1+x^2}\,dx$$

  • $\begingroup$ Wolfram Alpha doesn't find a closed-form solution to this. Are you certain that a closed-form answer exists? $\endgroup$ Dec 20 '13 at 6:02
  • 17
    $\begingroup$ I want a perfectly-baked Margherita pizza. $\endgroup$
    – dfeuer
    Dec 20 '13 at 6:03
  • 5
    $\begingroup$ Well, I want a pony, but unlike the OP, I don't seem to be getting one :( $\endgroup$
    – Igor Rivin
    Dec 20 '13 at 6:19
  • 1
    $\begingroup$ It is useful to know about inverse hyperbolic functions. $\endgroup$ Dec 20 '13 at 8:26
  • $\begingroup$ @arbautjc Thank you for your observation. I've removed my answer. $\endgroup$
    – Alraxite
    Dec 20 '13 at 14:58

Why not make the substitution $x=\sinh u?$ Then, $d x = \cosh u d u,$ while $(x+\sqrt{1+x^2})= \exp(u),$ and $1+x^2 = \cosh^2 u,$ so we get our integral equal to

$$\int \frac{\sqrt u}{\cosh u} du,$$ Now, letting $u = v^2,$ this is equal to $$\frac12\int \operatorname{sech}(v^2) d v.$$ At which point I am flummoxed.

  • $\begingroup$ @ronno oops, you are not wrong :( $\endgroup$
    – Igor Rivin
    Dec 20 '13 at 6:39
  • $\begingroup$ @ronno though this seems to make it easier... $\endgroup$
    – Igor Rivin
    Dec 20 '13 at 6:46
  • $\begingroup$ Maybe I'm missing something now, but how did you go from $\frac{2v^2}{\cosh(v^2)}dv$ to $\frac12\exp(-v^2) dv$? $\endgroup$
    – ronno
    Dec 20 '13 at 6:53
  • $\begingroup$ @ronno by hallucinating (in the middle of the night). $\endgroup$
    – Igor Rivin
    Dec 20 '13 at 13:53
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    $\begingroup$ By the definition of sech there is going yo be a error functiony thing in there $\endgroup$ Dec 30 '13 at 21:00

I'd add this as a comment but I don't have enough reputation to do so. Building off Igor's work, I wondered whether $\frac{2v^2}{\cosh(v^2)}$ could be done by parts. If we let $f=v$ and $g'=2v\text{sech}(v^2)$, then $fg-\int{f'g}$ works out to $$2v\arctan(e^{v^2})-2\int{\arctan(e^{v^2})dv}$$

Of course, I had to go to Wolfram for the second half:

$$ 2 \int{\arctan(e^{v^2}) dv} = \sqrt{\frac{\pi}{\ln(\arctan(e))}}\text{erfi}\left(v\sqrt{\ln(\arctan(e))}\right)$$

where $\text{erfi}$ is apparently the imaginary error function. (I know a little about the plain ol' error function, but...) Unless there's something weird with that error function thing, I'd assume we can just undo the substitution, with $v=\sqrt{u}=\sqrt{\text{arcsinh}(x)}$:


I'm guessing, though, that most of the people looking at this know what they're doing more than I do, so I wouldn't be surprised if there's some reason we can't do the above.

I'm trying to check this on Wolfram, just by subtracting the original integrand from the derivative of the above, but it's not interpreting the parentheses correctly. This is what I'm entering:





It keeps taking the derivative of the whole thing. I'm pretty sure I parsed that carefully, but I thought I'd include it in case anyone else wants to try.

  • $\begingroup$ Wolfram Alpha integrated $\arctan(e)^{v^2}$ instead of $\arctan(e^{v^2})$. $\endgroup$
    – ronno
    Dec 21 '13 at 9:09
  • $\begingroup$ @ronno, hmm, it seems it did. I put the v^2 in parentheses; it didn't occur to me that e^(v^2) had to go in parentheses too. To the person who voted my response down: Could you please tell me why? I made a good-faith effort to answer the question; my error came not from sloppiness but from ignorance of how Wolfram parses input. An error made earlier by another member was one I spotted; it seemed possible I saw something others hadn't. I noted my uncertainties. I ask as I'm obviously trying, if anything, to preserve reputation. ronno's response would have sufficed for me to retract my answer. $\endgroup$
    – dmk
    Dec 21 '13 at 14:43

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