TITLE: Confusion in options of integration
QUESTION [5 upvotes]: If
  $$
I(a)=\int_0^\infty\frac{\arctan(ax)-\arctan(x)}{x}\,\mathrm dx,
$$
  then


(A) $I'(1),I'(2),I'(3)$ are in harmonic progression.
(B) $I'(2)=\dfrac\pi4$
(C) $I(\pi)=\dfrac\pi2\log\pi$
(D) $I'(3)=\dfrac\pi6$  

In this question the derivative of I should be zero as RHS of a $I$ is constant .
But the answer is given as $A,B,C,D$

REPLY [8 votes]: Note that $I(a)$ is a function of $a$.  Therefore, we can differentiate under the integral to obtain

$$\begin{align}
I'(a)&=\frac{d}{da}\int_0^\infty \frac{\arctan(ax)-\arctan(x)}{x}\,dx\\\\
&=\int_0^\infty \frac{\partial}{\partial a}\left(\frac{\arctan(ax)-\arctan(x)}{x}\right)\,dx\\\\
&=\int_0^\infty \frac{1}{1+(ax)^2}\,dx\tag 1\\\\
&=\frac{\pi}{2a}\tag 2
\end{align}$$
Then, using $I(1)=0$, we find after integrating $(2)$ that 
$$\bbox[5px,border:2px solid #C0A000]{I(a)=\frac{\pi \log(a)}{2}}\tag 3$$
Then, we see from $(2)$ and $(3)$ that all choices (A), (B), (C), and (D) are true.

NOTE:

The uniform convergence of the integral in $(1)$ for $|a|\ge \delta>0$ (coupled with the convergence of $I(a)$ for all $a\ne0$) guarantees the legitimacy of differentiating under the integral sign.