Skip to main content

$I = \int\limits_0^\infty {\ln \left( {\frac{{{e^x} + 1}}{{{e^x} - 1}}} \right)dx} $

(A) I > 2
(B) I < 1
(C) I = 2
(D) 1 < I < 2

$I = \int\limits_0^\infty  {\ln \left( {\frac{{1 + {e^{ - x}}}}{{1 - {e^{ - x}}}}} \right)dx} $

Let, ${e^{ - x}} = t$

So, $ - {e^{ - x}}dx = dt$

$\therefore dx =  - \frac{{dt}}{t}$

$I = \int\limits_1^0 {\ln \left( {\frac{{1 + t}}{{1 - t}}} \right)\left( { - \frac{1}{t}} \right)dt}  = \int\limits_0^1 {\frac{1}{t}\ln \left( {\frac{{1 + t}}{{1 - t}}} \right)dt} $

$ = \int\limits_0^1 {\frac{{\ln (1 + t)}}{t}} dt - \int\limits_0^1 {\frac{{\ln (1 - t)}}{t}dt} $

$ = \int\limits_0^1 {\frac{{t - \frac{{{t^2}}}{2} + \frac{{{t^3}}}{3} - \frac{{{t^4}}}{4} + ...........}}{t}} dt - \int\limits_0^1 {\frac{{ - t - \frac{{{t^2}}}{2} - \frac{{{t^3}}}{3} - \frac{{{t^4}}}{4} + ...........}}{t}} dt$

$ = \int\limits_0^1 {\left( {1 - \frac{t}{2} + \frac{{{t^2}}}{3} - \frac{{{t^3}}}{4} + ...........} \right)} dt - \int\limits_0^1 {\left( { - 1 - \frac{t}{2} - \frac{{{t^2}}}{3} - \frac{{{t^3}}}{4} - ...........} \right)} dt$

$ = \left. {\left( {t - \frac{{{t^2}}}{{{2^2}}} + \frac{{{t^3}}}{{{3^2}}} - \frac{{{t^4}}}{{{4^2}}} + .............} \right)} \right|_0^1 + \left. {\left( {t + \frac{{{t^2}}}{{{2^2}}} + \frac{{{t^3}}}{{{3^2}}} + \frac{{{t^4}}}{{{4^2}}} + .............} \right)} \right|_0^1$

$ = \left( {1 - \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} - \frac{1}{{{4^2}}} + ...........} \right) + \left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ............} \right)$

$ = 2\left( {1 + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{7^2}}}...........} \right)$

Hence, Option (A).

Popular posts from this blog