Question

The value of ${\int }_0^{\infty }\frac{dx}{1+x^4}$ is

• A. same as that of ${\int }_0^{\infty }\frac{x^2+1dx}{1+x^4}$
• B. $\frac{\pi }{2\sqrt{2}}$
• C. same as that of ${\int }_0^{\infty }\frac{x^{2}dx}{1+x^4}$
• D. $\frac{\pi }{\sqrt{2}}$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{\pi }{2\sqrt{2}}$
C same as that of ${\int }_0^{\infty }\frac{x^{2}dx}{1+x^4}$

$I={\int }_0^{\infty }\frac{dx}{1+x^4}$ ....(1)
${\int }_0^{\infty }\frac{x^2+1-x^2}{1+x^4}dx$
$={\int }_0^{\infty }\frac{x^2}{1+x^4}dx+{\int }_0^{\infty }\frac{1-x^2}{1+x^4}dx$ ....(2)
Let $I_1={\int }_0^{\infty }\frac{x^2}{1+x^4}dx$
and $I_2={\int }_0^{\infty }\frac{1-x^2}{1+x^4}dx$
$I_2={\int }_0^{\infty }\frac{\frac{1}{x^2}-1}{\frac{1}{x^2}+x^2}dx$
$I_2={\int }_0^{\infty }\frac{\frac{1}{x^2}-1}{\frac{1}{x^2}+x^2+2-2}dx$
$I_2={\int }_0^{\infty }\frac{\frac{1}{x^2}-1}{(x+\frac{1}{x}{)}^2-2}dx$

Put $x+\frac{1}{x}=y$
$\Rightarrow dy=(1-\frac{1}{x^2})dx$
Also, when $x=0\Rightarrow y=\infty$
And when $x=\infty \Rightarrow y=\infty$
$\therefore I_2={\int }_{\infty }^{\infty }\frac{-1}{y^2-2}dy=0$
$\therefore I={\int }_0^{\infty }\frac{x^{2}dx}{1+x^4}$ .....(3)
Adding equations (1) and (3), we get
$2I={\int }_0^{\infty }\frac{1+x^2}{1+x^4}dx$
$={\int }_0^{\infty }\frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2}dx$
$={\int }_0^{\infty }\frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2+2-2}dx$
$={\int }_0^{\infty }\frac{\frac{1}{x^2}+1}{(x-\frac{1}{x}{)}^2+2}dx$
Put $x-\frac{1}{x}=t$
$\Rightarrow dt=(1+\frac{1}{x^2})dx$

$2I={\int }_{-\infty }^{\infty }\frac{dt}{t^2+2}$
$=[\frac{1}{\sqrt{2}}ta{n}^{-1}\frac{t}{\sqrt{2}}{]}_{-\infty }^{\infty }=\frac{\pi }{\sqrt{2}}$
$\therefore I=\frac{\pi }{2\sqrt{2}}$