How can I compute the integral $\int_{0}^{\infty} \frac{dt}{1+t^4}$? I have to compute this integral $$\int_{0}^{\infty} \frac{dt}{1+t^4}$$ to solve a problem in a homework. I have tried in many ways, but I'm stuck. A search in the web reveals me that it can be do it by methods of complex analysis. But I have not taken this course yet. Thanks for any help.

Let the considered integral be I i.e
$$I=\int_0^{\infty} \frac{1}{1+t^4}\,dt$$
Under the transformation $t\mapsto 1/t$, the integral is:
$$I=\int_0^{\infty} \frac{t^2}{1+t^4}\,dt \Rightarrow 2I=\int_0^{\infty}\frac{1+t^2}{1+t^4}\,dt=\int_0^{\infty} \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\,dt$$
$$2I=\int_0^{\infty} \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2}\,dt$$
Next, use the substitution $t-1/t=u \Rightarrow (1+1/t^2)\,dt=du$ to get:
$$2I=\int_{-\infty}^{\infty} \frac{du}{u^2+2}\Rightarrow I=\int_0^{\infty} \frac{du}{u^2+2}=\boxed{\dfrac{\pi}{2\sqrt{2}}}$$
$\blacksquare$
