Draft:Ahmed's integral

Ahmed's integral is a definite integral over the unit interval which equals to $\frac{5{\pi }^2}{96}$ and is written as;

${\int }_0^1\left(\frac{𝕒𝕣𝕔𝕥𝕒𝕟(\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}\right)dx=\frac{5{\pi }^2}{96}$

The integral is used as a popular problem and puzzle in various webs and videos online. It was proposed by Zafar Ahmed during 2001 to 2002 in the American Mathematical Monthly.

Methods of solving

Substitution

One of the few ways of integrating this is by substitution.

Let the integral be $I$;

$I={\int }_0^1\left(\frac{𝕒𝕣𝕔𝕥𝕒𝕟(\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}\right)dx$

Then use $𝕒𝕣𝕔𝕥𝕒𝕟(z)=\frac{\pi }{2}-𝕒𝕣𝕔𝕥𝕒𝕟(\frac{1}{z})$ to split $I$ as $I=I_1-I_2$. Now substitute $x=𝕥𝕒𝕟(\theta )$;

$I_1=\frac{\pi }{2}{\int }_0^{\frac{\pi }{4}}\frac{𝕔𝕠𝕤(\theta )}{\sqrt{2-{𝕤𝕚𝕟}^2(\theta )}}d\theta$

Proceed by substituting $\sqrt{2}𝕤𝕚𝕟(\varphi )$ into $\theta$ which equates to;

$I_1=\frac{{\pi }^2}{12}$

Next, we can use the representation of;

$I=\frac{1}{a}𝕒𝕣𝕔𝕥𝕒𝕟\left(\frac{1}{a}\right){\int }_0^1\frac{1}{x^2+a^2}dx$, where $a\ne 0$

to express;

$I_2={\int }_0^1{\int }_0^1\frac{1}{(1+x^2)(2+x^2+y^2)}dxdy$.

Which can be rewritten as;

$I_2={\int }_0^1{\int }_0^1\frac{1}{(1+x^2)(1+y^2)}dxdy-{\int }_0^1{\int }_0^1\frac{1}{(1+y^2)(2+x^2+y^2)}dxdy$.

Which becomes;

$I_2=\frac{{\pi }^2}{32}$

And thus;

$I=\frac{{\pi }^2}{12}-\frac{{\pi }^2}{32}=\frac{5{\pi }^2}{96}$

Feynman's Trick

Another method is by using Feynman's Trick.

Begin with a 'u-parameterized' version of Ahmed's integral;

$I(u)={\int }_0^1\left(\frac{𝕒𝕣𝕔𝕥𝕒𝕟(u\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}\right)dx$

Differentiate it with respect to u. I(1) is Ahmed's integral. As u > inf, the argument for arctan also > inf for all x>0, since arctan(inf)=pi/2 then;

$I(\mathrm{\infty })=\frac{\pi }{2}{\int }_0^1\frac{1}{(1+x^2)\sqrt{2+x^2}}dx$