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

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Answer 1 · 2018-05-05T14:31:53

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

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} dx#.

Now, we may view #\sqrt{8} x# 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 =#

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

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

#= \frac{pi}{2 \sqrt{8}} =#

#= \frac{pi}{4 \sqrt{2}}#.