Show that the following integral: $$ \int_{-\infty}^{\infty} \frac{1}{x^4 + 1} \,dx = \frac{\pi}{\sqrt{2}}. $$ I know this is $\pi/\sqrt{2}$ which I was found with Mathematica. I think I could utilise the substitution $t = x^2 + 1$ or $t = 1/x$, but I can't seem to make an approach because the integral is quite different from what I have been dealing with. Is there a better approach?
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1See also: Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. – Ben Dec 22 '13 at 06:52
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This follows from the more general beta function approach given here: http://math.stackexchange.com/a/34761/6075 – Eric Naslund Dec 22 '13 at 06:56
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Your integral is a special case of the more general cases. – Mhenni Benghorbal Dec 22 '13 at 11:38
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HINT : $$x^4+1=(x^2+1)^2-2x^2=(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)$$
Then, find $A,B,C,D$ such that $$\frac{Ax+B}{x^2+\sqrt 2x+1}+\frac{Cx+D}{x^2-\sqrt 2x+1}.$$
mathlove
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I know a way that may work here. I am assuming the integrals is regarded as $$2\int_0^{\infty}\frac{dx}{1+x^4}$$ First, show that $$\int_0^{\infty}\frac{x^{p-1}}{1+x}dx=\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin(\pi p)},~~~0<p<1$$ This can be done by a especial substitution $y=\frac{x}{1+x}$. Second, after showing the above identity, then focus on $\int_0^\infty{}\frac{dx}{1+x^4}$. If we set $t=x^4$ then after doing some integration manipulation it becomes: $$\frac{1}4\int_0^{\infty}\frac{t^{-3/4}}{1+t}dt=\frac{\pi}{4\sin(\pi/4)}=\frac{\pi\sqrt{2}}{4}$$
Mikasa
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Hint: $x^4+1 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$. Then use partial fractions.
Bruno Joyal
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This can also be done with complex integration using half-circle contour going either in upper or lower half-plane. Complex part goes to zero when radius goes to infinity and integral is $2\pi i$ times sum of residues inside the contour.
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