Too Difficult?

Consider the integral $f(a)=\int^{\frac{\pi}{2}}_0\frac{\tan^{-1}(a\tan x)}{\tan x}dx$ Differentiating under the integral sign (with respect to $a$ ), we get $f'(a)=\int^\frac{\pi}{2}_0\frac{1}{a^2\tan^2x+1}dx=\frac{\pi}{2a+2}$ Now we take the antiderivative of it. $f(a)=\frac{\pi}{2}\ln|a+1|+c$ But we know that $f(0)=0$ , so we get $c=0$ .

From that, substituting, $a=1$ , we get $\boxed{\frac\pi2\ln2}$

First we see that the integrand is $x \cot x$ , and we can easily see that $\cot x \, dx = d(\log |\sin x|)$ . Therefore, the integral is $\int_{x=0}^{x=\pi/2} x \, d(\log |\sin x|).$ Note that we can abandon absolute since it takes only nonnegative values. Using integration by parts gives $\big[ x \log\sin x\big]_{x=0}^{x=\pi/2}-\int_{x=0}^{x=\pi/2} \log\sin x\, dx.$ For the first part, as $x \to \pi/2$ , we see that it tends to zero. And as $x \to 0^+$ , we find that the limit is of the indeterminate form $0 \cdot \infty$ . We express it as follow: $\lim_{x \to 0^+} \frac{\log\sin x}{1/x}.$ It is of the form $\infty/\infty$ , so we use L'Hopital rule and get $\lim_{x \to 0^+} \frac{\frac{\cos x}{\sin x}}{-1/x^2} \, = \, \lim_{x \to 0^+} -(x \cos x)\left(\frac{x}{\sin x}\right)=0.$ Therefore, the first part is exactly zero, thus we now have $\int_{x=0}^{x=\pi/2} x \cot x \, dx \, = \, - \int_{x=0}^{x=\pi/2} \log\sin x\, dx.$ Define $A$ to be the term on right hand side, which is our goal. If we substitute by changing $x$ to be $\frac{\pi}{2}-x$ , we get that $A=-\int_{0}^{\pi/2} \log\cos x\, dx.$ Therefore, $2A=-\int_{0}^{\pi/2} \log\sin x+\log\cos x\, dx = -\int_{0}^{\pi/2} \log(\sin x \cos x), dx$ By the identity $\sin 2x=2\sin x \cos x$ , we have $2A=-\int_{0}^{\pi/2} -\log 2+\log(\sin 2x) \, dx = \frac{\pi}{2}\log 2-\int_{0}^{\pi/2} \log\sin 2x\, dx$ All we have left to show is that the right integral is actually $A$ itself. To show that, substitute $x$ by $\frac{x}{2}$ . We then have $-\int_{0}^{\pi/2} \log\sin 2x\, dx \,=\, -\int_{0}^{\pi} \frac{1}{2}\log \sin x \, dx.$ Notice that the function $\log \sin x$ is symmetry upon the axis $x=\pi/2$ . It means that the integral from $0$ to $\pi/2$ equals the integral from $\pi/2$ to $\pi$ . So we have $-\int_{0}^{\pi} \frac{1}{2}\log \sin x \, dx \,=\,-\int_{0}^{\pi/2} \log\sin x\, dx = A.$ This completes the proof that desired integral is $\frac{\pi}{2}\log 2$ , since we have $2A=\frac{\pi}{2}\log 2+A$ .

Let $u = \tan x$ . The integral then becomes

$\int_0^\infty \frac{\tan^{-1} u}{u(1+u^2)} \ du.$

We then introduce a variable a, and consider the integral

$I(a) = \int_0^\infty \frac{\tan^{-1}(au)}{u(1+u^2)} \ du$

Differentiating through the integral wrt a we get

$\frac{d}{da} I(a) = \int_0^\infty \frac{d}{da} \frac{\tan^{-1}(au)}{u(1+u^2)} \ du = \int_0^\infty \frac{1}{(1+a^2u^2)(1+u^2)} \ du,$

which may be evaluated using partial fractions. Using the cover-up method or otherwise, we find that $\dfrac{1}{(1+a^2u^2)(1+u^2)} = \dfrac{1/(1-a^2)}{1+u^2} - \dfrac{a^2/(1-a^2)}{1+a^2u^2}$ . Hence,

$\begin{aligned} \int_0^\infty \frac{1}{(1+a^2u^2)(1+u^2)} \ du &= \int_0^\infty \frac{1/(1-a^2)}{1+u^2} - \frac{a^2/(1-a^2)}{1+a^2u^2} \ du \\ &= \left[ \frac{\tan^{-1} u}{1-a^2} - \frac{a\tan^{-1}(au)}{1-a^2} \right]_0^\infty \\ &= \frac{\pi}{2} \frac{1}{1-a^2} - \frac{\pi}{2} \frac{a}{1-a^2} \\ &= \frac{\pi}{2(1+a)}. \end{aligned}$

Therefore,

$I(a) = \int_0^a \frac{\pi}{2(1+\alpha)} \ d\alpha = \frac{\pi}{2}\ln(1+a) + C,$

and C = 0 since I(0) = 0. From this finally we may finally conclude that $\displaystyle \int_0^\infty \frac{\tan^{-1} u}{u(1+u^2)} \ du = I(1) = \frac{\pi}{2}\ln 2$ .