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)?

ADDENDUM

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)

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

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