# What is the value of the following definite integral?

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

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

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

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

$$\therefore \, 4I=0$$

$$\therefore \, I=0$$

But DESMOS gives me the answer approaching $$2.53824756838$$

What wrong with my approach?

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

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):