2
$\begingroup$

$I=\displaystyle\int\limits_0^\infty \dfrac{1}{(1+x^2)\displaystyle\sqrt{\log(1+x)}}dx$

My attempt:

Substituting $x=\tan{\theta}$

$I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{1}{\displaystyle\sqrt{\log(1+\tan\theta)}}d\theta$

By applying $\,\,\int\limits_a^b f(x)dx=\int\limits_a^b f(a+b-x)dx$

$I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{1}{\displaystyle\sqrt{\log(1+\cot\theta)}}d\theta$

Adding both and simplifying,

$2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{\displaystyle\sqrt{\log(1+\cot\theta)}+\displaystyle\sqrt{\log(1+\tan\theta)}}{\displaystyle\sqrt{\log(1+\cot\theta)\log(1+\tan\theta)}}d\theta$

Rationalizing the numerator,

$2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{\log(1+\cot\theta)-\log(1+\tan\theta)}{\displaystyle\sqrt{\log(1+\cot\theta)\log(1+\tan\theta)}\displaystyle\left(\sqrt{\log(1+\cot\theta)}-\displaystyle\sqrt{\log(1+\tan\theta)}\right)}d\theta$

$\therefore\, 2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{\log(1+\tan\theta)-\log(1+\tan\theta)-\log\tan\theta}{\displaystyle\sqrt{\log(1+\cot\theta)\log(1+\tan\theta)}\displaystyle\left(\sqrt{\log(1+\cot\theta)}-\displaystyle\sqrt{\log(1+\tan\theta)}\right)}d\theta$

$\therefore\, 2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{-\log\tan\theta}{\displaystyle\sqrt{\log(1+\cot\theta)\log(1+\tan\theta)}\displaystyle\left(\sqrt{\log(1+\cot\theta)}-\displaystyle\sqrt{\log(1+\tan\theta)}\right)}d\theta$

Again applying the property $\,\,\int\limits_a^b f(x)dx=\int\limits_a^b f(a+b-x)dx$

$\therefore\, 2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{-\log\cot\theta}{\displaystyle\sqrt{\log(1+\tan\theta)\log(1+\cot\theta)}\displaystyle\left(\sqrt{\log(1+\tan\theta)}-\displaystyle\sqrt{\log(1+\cot\theta)}\right)}d\theta$

Adding both equations,

$ 2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{-\log\tan\theta+\log\tan\theta}{\displaystyle\sqrt{\log(1+\tan\theta)\log(1+\cot\theta)}\displaystyle\left(\sqrt{\log(1+\tan\theta)}-\displaystyle\sqrt{\log(1+\cot\theta)}\right)}d\theta$

$\therefore \, 4I=0$

$\therefore \, I=0$

But DESMOS gives me the answer approaching $2.53824756838$

What wrong with my approach?

$\endgroup$
  • $\begingroup$ The sign of $\log \tan$ at the numerator seems wrong after the last addition (don't know if everything was fine before). $\endgroup$ – N74 Oct 13 '18 at 11:49
  • $\begingroup$ All else as given, the numerator after the last addition (which btw is already $4I$ not $2I$) should be $-\log \cot\theta + \log\tan\theta$ $\endgroup$ – Lee David Chung Lin Oct 13 '18 at 17:37
  • $\begingroup$ A more accurate value (according to Mathematica) would be $2.538250393139 \ldots$ $\endgroup$ – Yuriy S Oct 13 '18 at 23:59
3
$\begingroup$

Doubt this integral has an elementary form, however it can be expressed as a series:

$$\int_0^\infty \frac{dx}{(1+x^2) \sqrt{\ln(1+x)}}=\int_0^\infty \frac{e^u du}{(1+(e^u-1)^2) \sqrt{u}}=2\int_0^\infty \frac{e^{v^2} dv}{2e^{-v^2}-2+e^{v^2}}$$

We work with the last integral:

$$2\int_0^\infty \frac{e^{v^2} dv}{2e^{-v^2}-2+e^{v^2}}=2\int_0^\infty \frac{e^{-v^2}dv}{1-2e^{-v^2}(1-e^{-v^2})}$$

It's easy to check that $|2e^{-v^2}(1-e^{-v^2})|<1$, so we can use the geometric series formula:

$$2\int_0^\infty \frac{e^{-v^2}dv}{1-2e^{-v^2}(1-e^{-v^2})}=2\sum_{n=0}^\infty 2^n \int_0^\infty e^{-(n+1)v^2}(1-e^{-v^2})^n dv=$$

Now we use the binomial sum formula:

$$=2\sum_{n=0}^\infty 2^n \sum_{k=0}^n (-1)^k \binom{n}{k} \int_0^\infty e^{-(n+k+1)v^2}dv=\sqrt{\pi} \sum_{n=0}^\infty 2^n \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{\sqrt{n+k+1}}$$

For the last part we used the well known Poisson integral formula. Finally:

$$\int_0^\infty \frac{dx}{(1+x^2) \sqrt{\ln(1+x)}}=\sqrt{\pi} \sum_{n=0}^\infty 2^n \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{\sqrt{n+k+1}}$$

It is easy to check numerically with Mathematica that the series gives the same value as the integral.

At a first glance, the convergence of this series is doubtful. I will not prove it here, but numerically with Mathematica it's clear that the series converges, and fast (here's the sum plotted vs the number of terms):

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.