- #1
Euge
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Evaluate the definite integral $$\int_0^\infty \frac{x^2 + 1}{x^4 + 1}\, dx$$
Using ##(2n)!! = 2n (2n-2) \cdots 2 = 2^n n!##,anuttarasammyak said:Another approach:
transforming x by
[tex]x=\tan\theta[/tex]
The integral is
[tex]\int_0^\pi \frac{d\phi}{1+\cos^2 \phi}=\sum_{n=0}^\infty (-1)^n \int_0^\pi \cos^{2n}\phi d\phi=\pi \sum_{n=0}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!}[/tex]
where
[tex]\phi=2\theta[/tex]
The series converges slowly. The sum from n=0 upto 50,000 gives the estimation between 0.7058
and 0.7083 for the expected ##1/\sqrt{2}##=0.70710...
julian said:Method #1
Using ##u = x^4##,
\begin{align*}
\int_0^\infty \dfrac{x^2+1}{x^4+1} dx = \frac{1}{4} \int_0^\infty \dfrac{u^{-1/4}+u^{-3/4}}{u+1} du \qquad (*)
\end{align*}
Here:
www.physicsforums.com/threads/exponential-type-integrals.1046843/
in post #12 I proved that, if ##0< a < 1##, then
$$
\int_0^\infty \frac{u^{a-1}}{1+u} du = \frac{\pi}{\sin a \pi}
$$
Using this in ##(*)##, we have
\begin{align*}
\int_0^\infty \dfrac{x^2+1}{x^4+1} dx & = \frac{\pi}{4} \left( \frac{1}{\sin 3 \pi / 4} + \frac{1}{\sin \pi / 4} \right)
\nonumber \\
& = \frac{\pi}{\sqrt{2}}
\end{align*}
Yes, you get out the additional result:pasmith said:In method #1, doing the contour integration directly also gets the value of [tex]
\int_0^\infty \frac{1 - x^2}{1 + x^4}\,dx.[/tex]