How to calculate $$\int_{0}^{\infty}\frac{\cos(mx)}{1+x^{4}}\,dx$$ where $m$ is a constant?
Any hint will be appreciated
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1Have you tried with contour integration? – Crostul Nov 07 '16 at 08:11
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look at this answer for the special case $m = 1$. – achille hui Nov 07 '16 at 08:29
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Contour integration along a proper chosen half-circle after reducing $\int_0^\infty=\frac12\int_{-\infty}^\infty$. The answer. – A.Γ. Nov 07 '16 at 09:05
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OK，thanks a lot – littleZ Nov 07 '16 at 11:05
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By the residue theorem, $$ \int_{0}^{+\infty}\frac{\cos(mx)}{1+x^4}\,dx = \frac{\pi}{2\sqrt{2}}e^{-|m|/\sqrt{2}}\left(\sin\frac{|m|}{\sqrt{2}}+\cos\frac{|m|}{\sqrt{2}}\right).$$
Jack D'Aurizio
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Per $\int_{0}^{\infty}\frac{\cos(m x)}{a^2+x^2}\,dx = \frac\pi{2a} e^{-m a}$, along with $a=e^{-i\frac\pi4 }$ $$ \int_{0}^{\infty}\frac{\cos(m x)}{x^4+1}\,dx = \Im \int_{0}^{\infty}\frac{\cos(m x)}{x^2+a^2}\,dx\\ =\frac\pi2 \Im \ e^{i\frac\pi4 } e^{-m e^{-i\frac\pi4 }} = \frac{\pi}{2}e^{-\frac m{\sqrt2}}\sin\left(\frac{m}{\sqrt{2}}+\frac\pi4 \right) $$
Quanto
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