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;

$\int _{0}^{1}{\Biggl (}{\frac {\mathbb {arctan} ({\sqrt {2+x^{2}}})}{(1+x^{2}){\sqrt {2+x^{2}}}}}{\Biggr )}dx={\frac {5\pi ^{2}}{96}}$

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 $I$;

$I=\int _{0}^{1}{\Biggl (}{\frac {\mathbb {arctan} ({\sqrt {2+x^{2}}})}{(1+x^{2}){\sqrt {2+x^{2}}}}}{\Biggr )}dx$

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

$I_{1}={\frac {\pi }{2}}\int _{0}^{\frac {\pi }{4}}{\frac {\mathbb {cos} (\theta )}{\sqrt {2-\mathbb {sin} ^{2}(\theta )}}}d\theta$

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

$I_{1}={\frac {\pi ^{2}}{12}}$

Next, we can use the representation of;

$I={\frac {1}{a}}\mathbb {arctan} {\biggl (}{\frac {1}{a}}{\biggr )}\int _{0}^{1}{\frac {1}{x^{2}+a^{2}}}\,dx$, where $a\neq 0$

to express;

$I_{2}=\int _{0}^{1}\int _{0}^{1}{\frac {1}{(1+x^{2})(2+x^{2}+y^{2})}}\,dx\,dy$.

Which can be rewritten as;

$I_{2}=\int _{0}^{1}\int _{0}^{1}{\frac {1}{(1+x^{2})(1+y^{2})}}\,dx\,dy-\int _{0}^{1}\int _{0}^{1}{\frac {1}{(1+y^{2})(2+x^{2}+y^{2})}}\,dx\,dy$.

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;

$I(u)=\int _{0}^{1}{\Biggl (}{\frac {\mathbb {arctan} (u{\sqrt {2+x^{2}}})}{(1+x^{2}){\sqrt {2+x^{2}}}}}{\Biggr )}dx$

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;

$I(\infty )={\frac {\pi }{2}}\int _{0}^{1}{\frac {1}{(1+x^{2}){\sqrt {2+x^{2}}}}}dx$