integration
Using contour integration I found that
$\int_0^{\infty} \frac{\cos{m x}}{1+x^4} dx = \frac{\pi}{2\sqrt{2}} e^{-\frac{|m|}{\sqrt{2}}} (\cos\frac{m}{\sqrt{2}} + \sin\frac{|m|}{\sqrt{2}})$
Is there any way to evaluate the integral using the differentiation under the integral sign, or any other technique that does not require contour integrals?
Aside from some trigonometric substitutions and identities, we will need the following identity, which can be shown using integration by parts twice: $$ \int_0^{\infty}\cos(\alpha t)e^{-\lambda t}\,\mathrm{d}t=\frac{\lambda}{\alpha^2+\lambda^2}\tag{1} $$ We will also use the standard arctangent integral: $$ \int_0^\infty\frac{\mathrm{d}t}{a^2+t^2}=\frac\pi{2a}\tag{2} $$ Now $$ \begin{align} &\left(\int_0^\infty\color{#C00000}{\sin}(x^2) e^{-\lambda x^2}\,\mathrm{d}x\right)^2\\ &=\int_0^\infty\int_0^\infty \color{#C00000}{\sin}(x^2)\color{#C00000}{\sin}(y^2) e^{-\lambda(x^2+y^2)}\,\mathrm{d}y\,\mathrm{d}x\tag{3.1}\\ &=\frac12\int_0^\infty\int_0^\infty \left(\cos(x^2-y^2) \color{#FF0000}{-}\cos(x^2+y^2)\right) e^{-\lambda(x^2+y^2)}\,\mathrm{d}y\,\mathrm{d}x\tag{3.2}\\ &=\frac12\int_0^{\pi/2}\int_0^\infty \left(\cos(r^2\cos(2\phi)) \color{#FF0000}{-}\cos(r^2)\right)e^{-\lambda r^2} \,r\,\mathrm{d}r\,\mathrm{d}\phi\tag{3.3}\\ &=\frac14\int_0^{\pi/2}\int_0^\infty \left(\cos(s\cos(2\phi)) \color{#FF0000}{-}\cos(s)\right) e^{-\lambda s} \,\mathrm{d}s\,\mathrm{d}\phi\tag{3.4}\\ &=\frac14\int_0^{\pi/2}\left(\frac{\lambda}{\cos^2(2\phi)+\lambda^2} \color{#FF0000}{-}\frac{\lambda}{1+\lambda^2}\right)\,\mathrm{d}\phi\tag{3.5}\\ &=\frac12\int_0^{\pi/4} \frac{\lambda}{\cos^2(2\phi)+\lambda^2}\,\mathrm{d}\phi \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.6}\\ &=\frac14\int_0^{\pi/4} \frac{\lambda\,\mathrm{d}\tan(2\phi)} {1+\lambda^2+\lambda^2\tan^2(2\phi)} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.7}\\ &=\frac14\int_0^\infty\frac{\mathrm{d}t}{1+\lambda^2+t^2} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.8}\\ &=\frac{\pi/8}{\sqrt{1+\lambda^2}} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.9} \end{align} $$
$(3.1)$ change the square of the integral into a double integral
$(3.2)$ use $2\color{#C00000}{\sin}(A)\color{#C00000}{\sin}(B)=\cos(A-B)\color{#FF0000}{-}\cos(A+B)$
$(3.3)$ convert to polar coordinates
$(3.4)$ substitute $s=r^2$
$(3.5)$ apply $(1)$
$(3.6)$ pull out the constant and apply symmetry to reduce the domain of integration
$(3.7)$ multiply numerator and denominator by $\sec^2(2\phi)$
$(3.8)$ substitute $t=\lambda\tan(2\phi)$
$(3.9)$ apply $(2)$
Finally, take the square root of both sides of $(3)$ and let $\lambda\to0^+$ to get $$ \int_0^\infty\color{#C00000}{\sin}(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{4} $$
Addendum
I just noticed that the same proof works for $$ \int_0^\infty\cos(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{5} $$ if each red $\color{#C00000}{\sin}$ is changed to $\cos$ and each red $\color{#FF0000}{-}$ sign is changed to $+$.
I figured out a rigorous proof by myself.
If $t = 0$, then it is an integration of a real-valued function, and clearly, $\int_0^\infty e^{-x} dx = 1$.
If $t \neq 0$, without losing of generality, assume $t > 0$. Consider the contour below:
In the picture, $n$ is a positive number that will be sent to $\infty$, the top line passes the origin and the point $(-1, t)$. And we set $f(z) = e^z, z \in \mathbb{C}$. By Cauchy's integration theorem, \begin{align} 0 = \int_\Gamma f(z) dz = \int_{\Gamma_1} e^z dz + \int_{\Gamma_2} e^z dz + \int_{\Gamma_3} e^z dz \end{align}
Let's denote the angle between $\Gamma_1$ and the real axis by $\theta_0$.
Clearly, $\int_{\Gamma_3}e^z dz = \int_{-n}^0 e^x dx = 1 - e^{-n}$.
On $\Gamma_1$, $z$ has the representation $z = (it - 1)x, x \in (0, n/|1 - it|)$, thus \begin{align*} \int_{\Gamma_1}e^z dz = \int_0^{n/\sqrt{1 + t^2}}e^{(it - 1)x}(it - 1) dx = (it - 1) \int_0^{n/\sqrt{1 + t^2}} e^{(it - 1)x} dx \end{align*}
To get the desired result, it remains to show $\int_{\Gamma_2} e^z dz \to 0$ as $n \to \infty$. Let $z = ne^{i\theta}$ with $\theta \in (\theta_0, \pi)$. It follows that \begin{align*} & \left|\int_{\Gamma_2} e^z dz\right| = \left|\int_{\theta_0}^\pi e^{ne^{i\theta}}nie^{i\theta} d\theta\right| \\ \leq & \int_{\theta_0}^\pi e^{n\cos\theta}n d\theta \\ \leq & ne^{n\cos{\theta_0}}(\pi - \theta_0) \to 0 \end{align*} as $n \to \infty$. Here we used the fact that $\pi/2 < \theta_0 < \pi$.
Best Answer
Let $I(a) = \int_0^\infty \frac{\sin at} {t(t^2+1)}dt $, which satisfies $$I(a)-I’’(a) = \int_0^\infty \frac{\sin at}t dt= \frac\pi2$$ and hence has the solution $I(a) = \frac\pi2 (1-e^{-a}) $. Then
\begin{align} \int_{0}^{\infty}\frac{\cos mx}{x^4+1} dx = &\ \frac1{4\sqrt2}\int_{-\infty}^{\infty} \overset{x=\frac{t-1}{\sqrt2}}{\frac{(\sqrt2+x)\cos mx}{x^2+\sqrt2x+1}} +\overset{x= \frac{t+1}{\sqrt2 }}{\frac{(\sqrt2-x)\cos mx}{x^2-\sqrt2x+1}} \ d x\\ =& \ \frac1{\sqrt2}\int_{0}^{\infty} \frac{\cos \frac {mt }{\sqrt2} \cos\frac m{\sqrt2}}{t^2+1} + \frac{t\sin\frac {|m|t}{\sqrt2} \sin\frac {|m|}{\sqrt2}}{t^2+1}\ dt \\ =& \ \frac1{\sqrt2}\cos\frac{m}{\sqrt2}\cdot I’(\frac {m}{\sqrt2}) -\frac1{\sqrt2}\sin\frac{|m|}{\sqrt2}\cdot I’’(\frac {|m|}{\sqrt2})\\ = &\ \frac{\pi}{2\sqrt{2}}e^{-\frac{|m|}{\sqrt{2}}}\left(\cos\frac{m}{\sqrt{2}}+\sin\frac{|m|}{\sqrt{2}}\right) \end{align}