Compute $\int_0^{+\infty} \frac{e^{-t}}{t^4+1}dt$

Question

Problem:

Compute $\int_0^{+\infty} \frac{e^{-t}}{t^4+1}dt$ possibly using Residue Theorem. I got this problem solving and understanding the following:

Compute $\int_0^{+\infty}\frac{\sin x + \cos x}{x^4+1}dx$

My attempt:

I tried defining $f(z)=\frac{e^{-z}}{z^4+1}$ and integrating in the first quadrant in an opportune slice, but I cannot reach any results. Is there a way to compute this integral?

Answer 1

Without complex analysis or residues, you can compute the antiderivative.

Use the roots of unity and then partial fraction decomposition to face four integrals looking like $$I_a=\int \frac {e^{-t}}{t+a}$$ where $a$ is a complex number. This makes $$I_a=e^a\, \text{Ei}(-(t+a))$$

For sure, the final result is not very pretty $$\int_0^{+\infty} \frac{e^{-t}}{t^4+1}dt=\frac{1}{4 \sqrt{2} \pi ^{3/2}}\,\,G_{1,5}^{5,1}\left(\frac{1}{256}\right.\left| \begin{array}{c} \frac{3}{4} \\ 0,\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{3}{4} \end{array} \right)$$ where appears the Meijer G function.

Numerically, this is $0.63047783491849835735$ which is not recognized by inverse symbolic calculators.

Answer 2

Maple evaluates this in terms of the exponential integral function $$ \operatorname{Ei}_1(z) := \int_1^\infty\frac{e^{-zt}}{t}\;dt $$ like this. Define $$ F(u) := \frac{u}{4}\;e^u\;\operatorname{Ei}_1(u) . $$ Then $$ \int_0^1\frac{e^{-t}}{t^4+1}\;dt = \sum F(u) $$ where the sum is over all solutions $u$ of $u^4+1 = 0$.
That is: $$ \int_0^1\frac{e^{-t}}{t^4+1}\;dt = F\left(\frac{1+i}{\sqrt{2}}\right) +F\left(\frac{-1+i}{\sqrt{2}}\right) +F\left(\frac{-1-i}{\sqrt{2}}\right) +F\left(\frac{1-i}{\sqrt{2}}\right) . $$

In my opinion, this is more understandable than the Meijer G function.