Question

If \( I =\int_{0}^{\infty}\left[\tan ^{-1}\left(\frac{\ln x }{1+(\ln x )^{2}}\right)\right] dx \), then the value of \( I +1 \) is (where [.] is G.I.F.)

Answer 1

\(\because\) ln(x) < 1+(lnx)² \(\forall\) x \(\in\) (0, \(\infty\))

Also \(-1<\frac{lnx}{1+(lnx)^2}<1\) 

⇒ tan^-1(-1) < tan^-1\((\frac{lnx}{1+(lnx)^2})<tan^{-1}1\)

(\(\because\) tan^-1 x is strictly increasing function)

⇒ \(\frac{-\pi}4<tan^{-1}(\frac{lnx}{1+(lnx)^2})<\frac{\pi}4\) (\(\because tan^{-1} 1 = \frac{\pi}4\) )

⇒ [tan^-1\((\frac{lnx}{1+(lnx)^2})=0\,or -1\)]

for x = 1, lnx = 0

⇒ tan^-1(\(\frac{lnx}{1+(lnx)^2}\)) = tan^-1 = 0

\([tan^{-1}(\frac{lnx}{1+(lnx)^2})]\)\(=\begin{cases}-1&;0<x<1\\ 0&;1\leq x\leq\infty \end{cases}\) where [.] denotes greatest integer function.

So, I = \(\int\limits_0^{\infty}[tan^{-1}(\frac{lnx}{1+(lnx)^2})]dx\) = \(\int\limits_0^1-1dx+\int\limits_1^\infty0dx\)

= \(-[\frac{x^2}2]_0^1+0\) 

 = \(\frac{-1}2\) 

\(\therefore\) I + 1 = \(\frac{-1}2+1 = \frac12\)