Question

If $f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{{\tan }^{-1}\left(t\right)}{t}dt\forall x\in R^{+}$, and the value of $f\left(e^2\right)-f\left(\frac{1}{e^2}\right)=\frac{k\pi }{2}$, then $k=$

Answer 1

$f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{{\tan }^{-1}\left(t\right)}{t}dt\therefore f\left(\frac{1}{x}\right)=\underset{1}{\overset{1/x}{\int }}\frac{{\tan }^{-1}\left(t\right)}{t}dt$

Put $t=\frac{1}{u}\Rightarrow dt=-\frac{1}{u^2}du$
$\therefore f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{{\tan }^{-1}\left(\frac{1}{u}\right)}{\frac{1}{u}}(-\frac{1}{u^2})du\Rightarrow f\left(\frac{1}{x}\right)=-\underset{1}{\overset{x}{\int }}\frac{{\tan }^{-1}\left(\frac{1}{u}\right)}{u}du\Rightarrow f\left(\frac{1}{x}\right)=-\underset{1}{\overset{x}{\int }}\frac{{\cot }^{-1}\left(u\right)}{u}du\Rightarrow f\left(\frac{1}{x}\right)=-\underset{1}{\overset{x}{\int }}\frac{{\cot }^{-1}\left(t\right)}{t}dt$
Now,
$f\left(x\right)-f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{{\cot }^{-1}\left(t\right)+{\tan }^{-1}t}{t}dt=\underset{1}{\overset{x}{\int }}\frac{\pi }{2}\times \frac{1}{t}dt=\frac{\pi }{2}\log x\Rightarrow f\left(e^2\right)-f\left(\frac{1}{e^2}\right)=\frac{\pi }{2}{\log }_e{e}^2=\pi \Rightarrow \frac{k\pi }{2}=\pi \Rightarrow k=2$