How to evaluate the following integral?
$$ \int _{-\infty }^{\infty }\!{\frac {\cos \left( x \right) }{{x}^{4}+1}}{dx} $$
Unlike this example, according to maple, the solution does not contain sine and cosine integrals. But how does it eavluate this kind of integrals? The method?
This can be done using residues, with the function $f(z) = \exp(i z)/(z^4 + 1)$ and a contour that goes along the real axis and returns on a circular arc in the upper half plane.
$$1.5442760\;\approx\;\pi\exp{\left( -\frac{\sqrt{2}}{2}\right)} \sin\left(\frac{\pi}{4} + \frac{\sqrt{2}}{2}\right) \;=\;\int_\mathbb{R}\; \frac{\cos(x)}{x^4+1}\, dx$$ holds true, but it is not the answer to the question.
Furthermore, I do not know, how Maple is doing to get it.
Generalising the proposed integral, let's proceed to determine $$I_p\:=\:\int_\mathbb{R}\; \frac{\cos x}{x^p+1}\, dx \qquad\text{where $p\in\mathbb{N}$ is even}.$$
Consider $f(x)=1/(x^p+1)$. Using residues we are computing the Fourier transform \begin{align} \hat f(\omega)\: & =\;\int_\mathbb{R}\; \frac{e^{-i\omega x}}{x^p+1}\, dx \\[2.7ex] & =\;\int_\mathbb{R}\; \frac{\cos(\omega x)}{x^p+1}\, dx \; -\;i\int_\mathbb{R}\; \frac{\sin(\omega x)}{x^p+1}\, dx \end{align} which is purely real since the last integral vanishes, its integrand $\frac{\sin(\omega x)}{x^p+1}$ being an odd function of $x$ for every $\omega$. And it's immediate that $\,\hat f\,$ is an even function then.
Once $\,\hat f(\omega)\,$ is made explicit the sought values can be obtained via evaluation: $I_p=\hat f(1)$.
Jordan's lemma can be applied
to calculate $\,\hat f(\omega)$, by assuming $\omega\le 0$ and with the real axis and the (infinite) semicircular arc in the upper half plane as contour of integration.
How does the integrand behave on that arc, parametrised by
$Re^{i\varphi}$ with $R\gg 0$ and $0\le\varphi\le\pi\:$?
$$\left| \frac{e^{-i\omega R(\cos\varphi + i\sin\varphi)}}
{(Re^{i\varphi})^p +1} \right|
\;=\; \frac{\left| e^{\omega R\sin\varphi} \right|}
{\left| {R^p e^{ip\varphi} +1} \right|}$$
The numerator is bounded above by $1$, and because of $p\ge 2$ the overall decrease is sufficient.
Let's turn to the residues
contributing to the value of the integral. They originate from the
$p$-th roots of $\,-1\,$ having positive imaginary part:
\begin{align}
z_k & \;=\;\exp\left(i(2k-1)\frac{\pi}{p}\right)
\quad\text{where }k=1,\ldots ,\frac{p}{2} \\
& \;=\;\cos\theta_k + i\sin\theta_k,\quad\theta_k=(2k-1)\frac{\pi}{p}
\end{align}
Note the reflection symmetry with respect to the imaginary axis which we record for later need $$\sin\theta_{\frac{p}{2}+1-k} = \sin\theta_k \quad\text{and}\quad \cos\theta_{\frac{p}{2}+1-k} = - \cos\theta_k .$$
Since $z_k$ is a simple root and the numerator $e^{-i\omega x}$ is non-zero, the corresponding residue is given by
\begin{align}
\operatorname{Res}(z_k) & \;=\; \frac{e^{-i\omega x}}{px^{p-1}}\Bigg|_{x=z_k}
\;=\; \frac{e^{-i\omega (\cos\theta_k +i\sin\theta_k)}}{pz_k^{p-1}} \\[2.7ex]
& \;=\; -\frac{1}{p} \exp(\omega \sin\theta_k)\,
e^{i\theta_k}e^{-i\omega\cos\theta_k}
\end{align}
When summing these up
from $\,k=1\,$ to $\,p/2\,$ the real parts will cancel pairwise due to the reflection symmetry$-$just "Go back to the roots!" If $\,p/2\,$ is odd, then $\,i\,$ belongs to the relevant root set, and the real part of that summand is zero anyway. After multiplying with $\,2\pi i\,$ we are done:
$$\hat f(\omega)\;=\;\frac{2\pi}{p}\sum_{k=1}^{p/2}\exp\big(-|\omega|\sin\theta_k\big)
\sin\big(\theta_k + |\omega|\cos\theta_k\big)$$
Until here $\omega\le 0$ was understood, to meet the conditions allowing for the calculus of residues. Upon introducing the absolute value of $\omega$, this assumption can be abandoned because $\hat f(\omega)$ is even.
Choosing $\,|\omega|=1$ we reach the goal initially set out:
$$I_p\;=\;\frac{2\pi}{p}\sum_{k=1}^{\frac{p}{2}} e^{-\sin\theta_k}\, \sin(\theta_k +\cos\theta_k)\, ,\quad\theta_k=(2k-1)\frac{\pi}{p}$$
Consider $f(z) = \exp(iz)/(z^4+1)$. Simple poles of $a(z)=\frac{b(z)}{c(z)}$ at $z_0$ have a residue of $\frac {b(z_0)}{c'(z_0)}$ at $z_0$. Thus, it can be seen that for any root $z_0$ of $z^4+1$, its residue is
$$\operatorname*{Res}_{z=z_0} f(z) = \frac{\exp(iz_0)}{4z_0^3}$$
Consider a semicircular contour of infinite radius, which, by Jordan's lemma, does not receive a contribution by the rounded part. There are two roots in this contour, $z_1 = (1+i)/\sqrt 2$ and $z_2 = (i-1)/\sqrt 2$. The residues at these poles are
$$r_1 = \operatorname*{Res}_{z = z_1}f(z) = \frac{\exp\left(\frac{i-1}{\sqrt 2}\right)}{(-2+2 i) \sqrt 2}$$
$$r_2 = \operatorname*{Res}_{z = z_2}f(z) = \frac{\exp\left(\frac{-1-i}{\sqrt 2}\right)}{(2+2 i) \sqrt 2}$$
Then,
$$I = \int_{-\infty}^\infty f(z)\,dz = 2\pi i (r_1+r_2) = \frac{\pi e^{-\frac{1}{\sqrt 2}} \left(\sin\left(\frac{1}{\sqrt 2}\right) + \cos\left(\frac{1}{\sqrt 2}\right)\right)}{\sqrt 2}$$
This is a two-sided variant of $$\int_0^\infty \frac{\cos(ax)}{b^4+x^4}dx=\frac{\pi\sqrt 2}{4b^3}\exp\left(-\frac{ab}{\surd 2}\right)\left[\cos\left(\frac{ab}{\surd 2}\right)+\sin\left(\frac{ab}{\surd 2}\right)\right].$$ My method is looking this up under item 3.727.1 in the Gradsteyn/Ryzhik tables.
\begin{align} &\int_{-\infty}^{\infty} \frac{\cos x}{x^{4}+1} d x \\ = &\ \frac1{2\sqrt2}\int_{-\infty}^{\infty} \overset{x=\frac{t-1}{\sqrt2}}{\frac{(x+\sqrt2)\cos x}{x^2+\sqrt2 x+1}} - \overset{x=\frac{t+1}{\sqrt2}}{\frac{(x-\sqrt2)\cos x}{x^2-\sqrt2 x+1}} \ d x\\ = &\ \frac1{\sqrt2}\int_{-\infty}^{\infty} \frac{\cos\frac t{\sqrt2} \cos\frac1{\sqrt2} + t\sin\frac t{\sqrt2} \sin\frac1{\sqrt2} }{t^2+1} \ d t \\ =& \ \frac1{\sqrt2}\left[I\bigg(\frac1{\sqrt2}\bigg)\cos\frac 1{\sqrt2}- I’\bigg(\frac1{\sqrt2}\bigg)\sin\frac 1{\sqrt2} \right]\\ = & \ \frac{\pi}{e^{\frac1{\sqrt2}}}\cos\bigg(\frac\pi4- \frac 1{\sqrt2}\bigg) \end{align} where $I(a) = \int_{-\infty}^\infty \frac{\cos at} {t^2+1}dt= \frac{\pi}{e^{a}}$ is assumed known.
