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.

You can also do it by partial fraction decomposition. We have
$$ \frac{2\sqrt{2}}{1+t^4} = \frac{t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{t - \sqrt{2}}{t^2 - \sqrt{2}t + 1}. $$
We have
$$\int_0^\infty \frac{dt}{t^2 \pm \sqrt{2}t + 1} =
\int_0^\infty \frac{dt}{(t \pm \sqrt{2}/2)^2 + 1/2} =
\int_0^\infty \frac{2dt}{(\sqrt{2} t \pm 1)^2 + 1} =
\sqrt{2} \arctan (\sqrt{2}t \pm 1) \big|_0^\infty.$$
Continuing, we get
$$\frac{\sqrt{2}\pi}{2} - \sqrt{2} \arctan (\pm 1) =
\frac{\sqrt{2}\pi}{2} \mp \frac{\sqrt{2}}{4} =
\frac{\sqrt{2}\pi (2 \mp 1)}{4}.$$
Next, we have
$$ \int_0^\infty \frac{(2t \pm \sqrt{2}) dt}{t^2 \pm \sqrt{2}t + 1} =
\log (t^2 \pm \sqrt{2}t + 1) \big|_0^\infty. $$
The integral doesn't converge, but we can consider instead
$$ \int_0^\infty \frac{(2t + \sqrt{2}) dt}{t^2 + \sqrt{2}t + 1} - \frac{(2t - \sqrt{2}) dt}{t^2 - \sqrt{2}t + 1} =
\log \frac{t^2 + \sqrt{2}t + 1}{t^2 - \sqrt{2}t + 1} \big|_0^\infty = 0. $$
Therefore we rewrite our initial expansion
$$ \frac{4\sqrt{2}}{1+t^4} = \frac{2t + 2\sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{2t - 2\sqrt{2}}{t^2 - \sqrt{2}t + 1} =
\frac{2t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{2t - \sqrt{2}}{t^2 - \sqrt{2}t + 1} + \frac{\sqrt{2}}{t^2 + \sqrt{2}t + 1} + \frac{\sqrt{2}}{t^2 - \sqrt{2}t + 1}. $$
Integrating, we get
$$ \int_0^\infty \frac{4\sqrt{2}}{1+t^4} =
\sqrt{2} \frac{\sqrt{2} \pi (2-1)}{4} + \sqrt{2} \frac{\sqrt{2} \pi (2+1)}{4} =
\frac{\pi}{2} + \frac{3\pi}{2} = 2\pi. $$
Therefore the integral we want is
$$ \int_0^\infty \frac{1}{1 + t^4} = \frac{2\pi}{4\sqrt{2}} = \frac{\sqrt{2}\pi}{4}. $$
