# Differentiating under the integral sign (differentiation with respect to a parameter)

In Example 3.3 of INTEGRATION: THE FEYNMAN WAY by an ANONYMOUS one finds also the same advice $G(t)=\displaystyle\int_{0}^{\pi /2}\dfrac{\arctan \left( t\tan x\right) }{\tan x}dx$.

I proceed as follows

$G^{\prime }(t)=\displaystyle\int_{0}^{\pi /2}\dfrac{\partial }{\partial t}\dfrac{\arctan\left( t\tan x\right) }{\tan x}dt=\int_{0}^{\pi /2}\dfrac{1}{t^{2}\tan ^{2}x+1}dx$

The difficult part now is how to integrate $\dfrac{1}{t^{2}\tan^{2}x+1}$ which is not detailed in the article.

After knowing the result, I worked backwards. By splitting the integrand into

$\dfrac{1}{t^{2}\tan ^{2}x+1}=\dfrac{t^{2}}{t^{2}-1}\dfrac{\tan ^{2}x+1}{t^{2}\tan ^{2}x+1}-\dfrac{1}{t^{2}-1}$

we can evaluate

$\displaystyle\int \dfrac{1}{t^{2}\tan ^{2}x+1}dx=\dfrac{t^{2}}{t^{2}-1}\int \dfrac{\tan^{2}x+1}{t^{2}\tan ^{2}x+1}dx-\int \dfrac{1}{t^{2}-1}dx$

$=\dfrac{t}{t^{2}-1}\arctan \left( t\tan x\right) -\dfrac{x}{t^{2}-1}$

Hence

$G^{\prime }(t)=\left. \dfrac{t\arctan \left( t\tan x\right) -x}{t^{2}-1}\right\vert _{0}^{\pi /2}$

$=\dfrac{t\arctan \left( t\tan \pi /2\right) -\pi /2}{t^{2}-1}-\dfrac{t\arctan\left( t\tan 0\right) -0}{t^{2}-1}$

$=\dfrac{\pi }{2\left( t+1\right) }$

and

$G(t)=\displaystyle\int \dfrac{\pi }{2\left( t+1\right) }\; dt=\dfrac{\pi }{2}\ln \left( t+1\right).$

(due to $G(0)=0$ the integration constant is $C=0$)

The original integral is therefore

$G(1)=\dfrac{\pi }{2}\ln 2.$

Does anyone know another way of evaluating $G(t)$?

Dear Todd

Thanks for reminding me that “the idea is to try to make the solutions mechanical”!

I should have thought of the standard substitution $x=\arctan u$. I checked the partial fractions technique you mentioned which gives

$\dfrac{1}{t^{2}u^{2}+1}\dfrac{1}{u^{2}+1}=\dfrac{t}{t^{2}-1}\dfrac{t}{t^{2}u^{2}+1}-\dfrac{1}{t^{2}-1}\dfrac{1}{u^{2}+1}.$

Of course the integral has the same value as before

$\displaystyle\int_{0}^{\infty }\frac{1}{t^{2}u^{2}+1}\frac{1}{u^{2}+1}du=\dfrac{t}{t^{2}-1}\displaystyle\int_{0}^{\infty }\dfrac{t}{t^{2}u^{2}+1}du-\dfrac{1}{t^{2}-1}\int_{0}^{\infty }\dfrac{1}{u^{2}+1}du$

$=\dfrac{t}{t^{2}-1}\left. \arctan \left( tu\right) \right\vert _{0}^{\infty}-\dfrac{1}{t^{2}-1}\left. \arctan \left( u\right) \right\vert _{0}^{\infty }$

$=\dfrac{t}{t^{2}-1}\left( \dfrac{\pi }{2}\right) -\dfrac{1}{t^{2}-1}\left( \dfrac{\pi }{2}\right)$

$=\dfrac{1}{t+1}\left( \dfrac{\pi }{2}\right)$

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