What is the improper integrals 1/((8x^2)+(1)) from 0 to infinity ?

1 Answer
May 5, 2018

int_0^\infty \frac{1}{8 x^2 + 1} dx = \frac{pi}{4 \sqrt{2}018x2+1dx=π42

Explanation:

To start with, we slightly rewrite the integral:

int_0^\infty \frac{1}{8 x^2 + 1} dx = int_0^\infty \frac{1}{( \sqrt{8} x )^2 + 1} dx018x2+1dx=01(8x)2+1dx.

Now, we may view \sqrt{8} x8x as an inner function, which is why this is the same as

\frac{1}{\sqrt{8}} int_0^\infty \frac{1}{y^2 + 1} dy =1801y2+1dy=

= \frac{1}{\sqrt{8}} [ tan^{-1} y ]_0^\infty ==18[tan1y]0=

= \frac{1}{\sqrt{8}} [ \frac{pi}{2} - 0 ] ==18[π20]=

= \frac{pi}{2 \sqrt{8}} ==π28=

= \frac{pi}{4 \sqrt{2}}=π42.