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?
$$ \begin{align} I(m) &:= \int_0^{\infty} \frac{\cos(m x)}{1+x^4} \mathrm{d}x = \frac12 \int_{-\infty}^{\infty} \frac{\cos(m x)}{1+x^4} \mathrm{d}x\\ \implies I'(m) &= -\frac12 \int_{-\infty}^{\infty} \frac{x\sin(m x)}{1+x^4} \mathrm{d}x\\ &= -\frac{1}{2\sqrt{8}} \int_{-\infty}^{\infty} \frac{\sin(m x)}{x^2-\sqrt{2}x+1} -\frac{\sin(m x)}{x^2+\sqrt{2}x+1}\mathrm{d}x\\ &= -\frac{1}{2\sqrt{8}} \int_{-\infty}^{\infty} \frac{\sin(m x)}{\left(x-\frac{1}{\sqrt{2}}\right)^2+\frac12} -\frac{\sin(m x)}{\left(x+\frac{1}{\sqrt{2}}\right)^2+\frac12}\mathrm{d}x\\ &= -\frac{1}{2\sqrt{8}} \int_{-\infty}^{\infty} \frac{\sin\left(m t +\frac{m}{\sqrt{2}}\right)}{t^2+\frac12} -\frac{\sin\left(m t -\frac{m}{\sqrt{2}}\right)}{t^2+\frac12}\mathrm{d}t\\ &= -\frac{\sin\left(\frac{m}{\sqrt{2}}\right)}{\sqrt{8}} \int_{-\infty}^{\infty} \frac{\cos(mt)}{t^2+\frac12}\mathrm{d}t\\ &= -\frac{\sin\left(\frac{m}{\sqrt{2}}\right)|m|}{\sqrt{8}} \int_{-\infty}^{\infty} \frac{\cos(s)}{s^2+\frac{m^2}{2}}\mathrm{d}s\\ &= -\frac{\pi}{2} \sin\left(\frac{m}{\sqrt{2}}\right)e^{-\frac{|m|}{\sqrt{2}}}\\ \implies I(m) &= \int_0^{m}-\frac{\pi}{2} \sin\left(\frac{x}{\sqrt{2}}\right)e^{-\frac{|x|}{\sqrt{2}}} \mathrm{d}x + I(0)\\ &=-\frac{\pi}{2}\frac{1-e^{-\frac{|m|}{\sqrt{2}}}\left(\cos\left(\frac{m}{\sqrt{2}}\right)+\sin\left(\frac{|m|}{\sqrt{2}}\right)\right)}{\sqrt{2}} + \int_0^{\infty} \frac{1}{1+x^4} \mathrm{d}x\\ &=\frac{\pi}{2\sqrt{2}}e^{-\frac{|m|}{\sqrt{2}}}\left(\cos\left(\frac{m}{\sqrt{2}}\right)+\sin\left(\frac{|m|}{\sqrt{2}}\right)\right), \end{align} $$ where the result needed to get from the seventh to the eighth line can be found here (and it's also derived without contour integration).
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}
Write $$x^4+1=(x-a)(x-b)(x-c)(x-d)$$ Use partial fraction decomposition $$\frac 1{x^4+1}=\frac A{x-a}+\frac B{x-b}+\frac C{x-c}+\frac D{x-d}$$ to face four integrals $$I(k)=\int \frac{\cos(x)}{x-k}\,dx=\int\frac{\cos (k+t)}{t}\,dt$$ Expand the cosine and use the fact that $k=\alpha+i\,\beta$ $$\cos (k+t)=\cos (\alpha ) \cosh (\beta ) \cos (t)-\sin (\alpha ) \cosh(\beta ) \sin (t)-$$ $$i (\cos (\alpha ) \sinh (\beta ) \sin (t)+\sin (\alpha ) \sinh (\beta ) \cos (t))$$ So, integrating, one sine integral and one cosine integral.
So, even if nasty, we have the antiderivative.
Back to $x$ and using the bounds $$\int_0^{\infty} \frac{\cos(x)}{x^4+1}\, dx=\frac{\pi }{2 \sqrt{2}}\,e^{-\frac{1}{\sqrt{2}}} \left(\sin \left(\frac{1}{\sqrt{2}}\right)+\cos \left(\frac{1}{\sqrt{2}}\right)\right)$$
No problem to generalize for any positive integer exponent in the denominator. The formula will be nasty for odd values of the exponent and nice for even values
$$\int_0^{\infty} \frac{\cos(x)}{x^2+1}\, dx=\frac{\pi }{2 e}$$ $$\int_0^{\infty} \frac{\cos(x)}{x^6+1}\, dx=\frac{\pi \left(1+\sqrt{3 e} \sin \left(\frac{\sqrt{3}}{2}\right)+\sqrt{e} \cos \left(\frac{\sqrt{3}}{2}\right)\right)}{6 e}$$
