I have to perform this definite integration:

$$I = \int_0^{+\infty}\ \text{arctanh}\left(\frac{1}{\sqrt{e^{1+x^2}}}\right)\ \text{d}x$$

I tried with a naive substitution

$$z = 1 + x^2 ~~~~~ \text{d}x = \frac{1}{2\sqrt{z-1}}$$


$$I = \int_1^{+\infty} \frac{\text{arctanh}(e^{-z/2})}{\sqrt{z-1}}\ \text{d}z$$

Which doesn't lean me anywhere. I tried with series expansion but the hyperbolic arctangent is hellish.

What I do know

According to Matematica $I = 0.825951$ but I have no clue how to proceed. Are there some useful substitutions/approximations for this kind of integral?

  • $\begingroup$ Do you mind me asking what is the motivation behind calculating this rather unnatural integral? $\endgroup$ – mickep Mar 2 '16 at 17:59
  • $\begingroup$ In your last formula, I believe the $e^{z/2}$ should be $e^{-z/2}$ $\endgroup$ – NoetherianCheese Mar 2 '16 at 18:06
  • $\begingroup$ @Cheese Right, I missed a minus whilst writing, thanks :D $\endgroup$ – Von Neumann Mar 2 '16 at 18:11

Integration by parts brings that integral into the slightly better form:

$$ I = \int_{0}^{+\infty} x^2 \frac{\exp\left(-\frac{1+x^2}{2}\right)}{1-\exp(-(1+x^2))}\,dx $$ and by expanding the integrand function as a geometric series, then computing $\int_{0}^{+\infty} x^2 \exp\left(-\frac{2k+1}{2}(1+x^2)\right)\,dx $ by usual means, we get:

$$ I = \sqrt{\frac{\pi}{2e}}\cdot\sum_{k\geq 0}\frac{1}{e^k (2k+1)^{3/2}} $$

where the last series is very fast-converging.

  • 1
    $\begingroup$ Uhhh! This is brilliant! $\endgroup$ – Von Neumann Mar 2 '16 at 18:10
  • 1
    $\begingroup$ @Jack, well done! (+1) The last sum is a Lerch transcendent, just in case u are interested! $\endgroup$ – tired Mar 2 '16 at 19:14

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