$$I=\int_0^{\infty}\frac{\cos x}{x^4+1}dx$$
I found this question where the denominator is simply $x+1$ but the answer there did not help. I'm assuming this requires contour integrals but I'm not very familiar with them. I am looking for any tips or methods to solve this. Thanks
Since $\frac{\cos x}{x^4+1}$ is even we can say: $$\int_0^{\infty}\frac{\cos x}{x^4+1}dx=\frac12\int_{-\infty}^{\infty}\frac{\cos x}{x^4+1}dx=\frac12\int_{-\infty}^{\infty}\frac{e^{ix}}{x^4+1}dx$$
Consider the contour integral of $\frac{e^{iz}}{z^4+1}$ along the path $L_R=[-R,R]$ and $C_R=\{|z|=R, Im(z)\geq0\}$, oriented counterclockwise. By Cauchy Integral Theorem we have: $$\int_{L_R}\frac{e^{iz}}{z^4+1}dz+\int_{C_R}\frac{e^{iz}}{z^4+1}dz$$ $$=\int_{\left| z-e^{\frac{\pi i}{4}} \right|=\frac12}\frac{e^{iz}}{z^4+1}dz+\int_{\left| z-e^{\frac{3\pi i}{4}} \right|=\frac12}\frac{e^{iz}}{z^4+1}dz$$ By the Cauchy Integral Formula: $$\int_{\left| z-e^{\frac{\pi i}{4}} \right|=\frac12}\frac{e^{iz}}{z^4+1}dz=\frac{2\pi ie^{\frac{i(\sqrt2+i\sqrt2)}{2}}}{\left(e^{\frac{\pi i}{4}}-e^{\frac{3\pi i}{4}}\right)\left(e^{\frac{\pi i}{4}}-e^{-\frac{\pi i}{4}}\right)\left(e^{\frac{\pi i}{4}}-e^{-\frac{3\pi i}{4}}\right)}$$ $$=\frac{\pi(1-i)e^{\frac{(-\sqrt2+i\sqrt2)}{2}}}{2\sqrt2}$$ Similarly: $$\int_{\left| z-e^{\frac{3\pi i}{4}} \right|=\frac12}\frac{e^{iz}}{z^4+1}dz=\frac{\pi(1+i)e^{\frac{(-\sqrt2-i\sqrt2)}{2}}}{2\sqrt2}$$ Therefore: $$\int_{L_R}\frac{e^{iz}}{z^4+1}dz+\int_{C_R}\frac{e^{iz}}{z^4+1}dz=\frac{\pi e^{-\frac{\sqrt2}{2}}}{\sqrt2}\left( \cos\frac{\sqrt2}{2}+\sin\frac{\sqrt2}{2} \right)$$ For $z$ laying on $C_R$, $y=Im(z)\geq0$ and therefore $|e^{iz}|=e^{-y}\leq1$. Hence: $$\left| \frac{e^{iz}}{z^4+1} \right|\leq\frac{1}{R^4-1}$$ so it follows that: $$\left| \int_{C_R}\frac{e^{iz}}{z^4+1}dz\right|\leq\frac{\pi R}{R^4-1}$$ Since $$\lim_{R\to\infty}\frac{\pi R}{R^4-1}=0$$ we can conclude that: $$\lim_{R\to\infty}\int_{C_R}\frac{e^{iz}}{z^4+1}dz=0$$ Therefore: $$ \begin{align*} \int_0^{\infty}\frac{\cos x}{x^4+1}dx=\frac12\int_{-\infty}^{\infty}\frac{e^{ix}}{x^4+1}dx \end{align*} \\ =\frac12\lim_{R\to\infty}\int_{L_R}\frac{e^{iz}}{z^4+1}dz \\ \color{blue}{=\frac{\pi e^{-\frac{\sqrt2}{2}}}{2\sqrt2}\left( \cos\left(\frac{\sqrt2}{2}\right)+\sin\left(\frac{\sqrt2}{2}\right) \right)} \\ $$
You can solve this using Feynman integration and distributions.
Let $$I(t)=\int_0^{\infty}\frac{\cos (tx)}{x^4+1}dx.$$
Then, $$I^{''''}(t)+I(t)=\int_0^{\infty}\cos (tx)dx.$$
Treating this integral as a distribution, you can write $\int_0^{\infty}\cos (tx)dx=\pi\delta(t)$ where $\delta(t)$ is the 'Dirac-delta function' which can be informally written as
$$\delta(x) = \begin{cases} +\infty & \text{if } x = 0, \\ 0 & \text{if } x \neq 0, \end{cases}$$
and always satifies $\int_a^b\delta(x)=1$ when $b<0$ and $a>0$. This condition is where the factor of $\pi$ comes from. Think about, for example, $\int_{-1}^{1}\left[\int_0^{\infty}\cos (tx)dx\right]dt=\int_0^{\infty}\int_{-1}^{1}\cos (tx)dtdx=\int_0^{\infty}\frac{\sin(t)}{t}dx=\pi$.
So, the ODE ends up as $$I^{''''}(t)+I(t)=\pi\delta(t).$$
This may look daunting but it's much more straightforward than you'd think. These kinds of equations are common in the study of Green's functions for ODE's.
To solve this in general, you solve $I^{''''}(t)+I(t)=0$ seperately for $t>0$ and $t<0$ and match up the coefficients so that $I(t)$ and its first and second derivatives are continuous at $0$ but $I^{'''}(0^+)-I^{'''}(0^-)=\pi$.
Now, we find need to find the initial values. We pick $0^+$. We can figure out $I$ and $I^{''}$ using partial fractions and the factorization $x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$. We can figure out $I'$ and $I^{'''}$ using the evenness of $I$ and the continuity conditions of $I$ at $0$.
The full initial value problem for $I$ is:
$$I^{''''}(t)+I(t)=\pi\delta(t);$$
$$I(0^+)=\frac{\pi}{2\sqrt{2}},\text{ }I'(0^+)=0,\text{ }I^{''}(0^+)=-\frac{\pi}{2\sqrt{2}},\text{ }I^{'''}(0^+)=\frac{\pi}{2}.$$ This has solution: $$I(t)=\frac{\pi e^{-\frac{|t|}{\sqrt{2}}} \left( \cos\left(\frac{t}{\sqrt{2}}\right) +\sin\left(\frac{|t|}{\sqrt{2}}\right) \right)}{2 \sqrt{2}}. $$
