Let f(x)= _0^xg(t) _e((1-t)/(1+t))dt, where g is a continuous odd function.

Question

Let $f(x)=\int _{0}^{x}g(t){\log }_{e}(\frac{1-t}{1+t})dt$, where $g$ is a continuous odd function.
If ${\int }_{\pi /2}^{-\pi /2}(f(x)+\frac{{x}^{2}\cos x}{1+{e}^{x}})dx={(\frac{\pi }{\alpha })}^{2}-\alpha $, then $\alpha $ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

$f(x)=\int _{0}^{x}g(t)\ln (\frac{1-t}{1+t})dt$

$f(-x)=\int _{0}^{-x}g(t)\ln (\frac{1-t}{1+t})dt$

$f(-x)=-\int _{0}^{x}g(-y)\ln (\frac{1+y}{1-y})dy$

$=-\int _{0}^{x}g(y)\ln (\frac{1-y}{1+y})dy$ (g is odd)

$f(-x)=-f(x)\Rightarrow f$ is also odd

Now,

$\begin{gathered} I=\int _{-\pi /2}^{\pi /2}(f(x)+\frac{{x}^{2}\cos x}{1+{e}^{x}})dx\, \ I=\int _{-\pi /2}^{\pi /2}(f(-x)+\frac{{x}^{2}{e}^{x}\cos x}{1+{e}^{x}})dx\, \ 2I=\int _{-\pi /2}^{\pi /2}{x}^{2}\cos xdx=2{\int }_{\pi /2}^{0}{x}^{2}\cos xdx \end{gathered}$

$\begin{gathered} I={({x}^{2}\sin x)}_{\pi /2}^{0}-\int _{0}^{\pi /2}2x\sin xdx \ =\frac{{\pi }^{2}}{4}-2{(-x\cos x+\int \cos xdx)}_{\pi /2}^{0} \ =\frac{{\pi }^{2}}{4}-2(0+1)=\frac{{\pi }^{2}}{4}-2\Rightarrow {(\frac{\pi }{2})}^{2}-2 \ 	herefore \alpha =2 \end{gathered}$

Solutions · 1