# definite integral #5

• Jun 16th 2011, 01:15 PM
Random Variable
definite integral #5
Show that $\displaystyle \int_{0}^{1} \frac{2a^{2}}{a^{2}+x^{2}} \frac{\arctan (\sqrt{2a^{2}+x^{2}})}{\sqrt{2a^{2}+x^{2}}} \ dx = \pi \arctan \Big(\frac{1}{\sqrt{2a^{2}+1}} \Big) - \Big( \arctan \frac{1}{a} \Big)^2$

If you let $\displaystyle a = 1$, you'll get a integral that maybe you've seen before.
• Jun 18th 2011, 01:52 PM
Random Variable
Re: definite integral #5
I'll offer a starting point.

Let p and q be polynomials. Then $\displaystyle \frac{1}{p(p+q)}} + \frac{1}{q(p+q)}} = \frac{1}{pq}$

Let $\displaystyle p = a^{2}+x^{2}$ and $\displaystyle q = a^{2}+t^{2}$

Then $\displaystyle \frac{1}{(a^{2}+x^{2})(2a^{2}+x^{2}+t^{2})} + \frac{1}{(a^{2}+t^{2})(2a^{2}+x^{2}+t^{2})} = \frac{1}{(a^{2}+x^{2})(a^{2}+t^{2})}$

Now integrate both sides over the unit square centered at $\displaystyle (\frac{1}{2}, \frac{1}{2})$

$\displaystyle \int^{1}_{0} \int^{1}_{0} \frac{dx \ dt}{(a^{2}+x^{2})(2a^{2}+x^{2}+t^{2})} + \int^{1}_{0} \int^{1}_{0} \frac{dx \ dt}{(a^{2}+t^{2})(2a^{2}+x^{2}+t^{2})} = \int^{1}_{0} \int_{0}^{1} \frac{dx \ dt}{(a^{2}+x^{2})(a^{2}+t^{2})}$

If you switch the order of integration of the second integral on the left, which is permissible since the integrand is never negative, you'll notice that the the two integrals on the left are identical.

$\displaystyle 2 \int^{1}_{0} \int^{1}_{0} \frac{dx \ dt}{(a^{2}+x^{2})(2a^{2}+ x^{2}+t^{2})} = \int^{1}_{0} \frac{dx}{a^{2}+x^{2}} \int_{0}^{1} \frac{dt}{a^{2}+t^{2}}$
• Jun 18th 2011, 04:00 PM
Krizalid
Re: definite integral #5
$\displaystyle \frac{\frac{\pi }{2}-\arctan \left( \sqrt{2{{a}^{2}}+{{x}^{2}}} \right)}{\sqrt{2{{a}^{2}}+{{x}^{2}}}}=\int_{0}^{1} {\frac{dt}{2{{a}^{2}}+{{x}^{2}}+{{t}^{2}}}},$ so we need to compute $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\frac{dx\,dt}{\left( {{a}^{2}}+{{x}^{2}} \right)\left( 2{{a}^{2}}+{{x}^{2}}+{{t}^{2}} \right)}}}.$