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The task is to calculate this integral using complex analysis: I know some methods, which involve calculating the integral around half of the circle, also it would be wise to add here $i cos(s^{\frac{1}{2}})$ and then take the real part of the result. But here comes my problem, this function isn't even, so I am unable to move the limit 0 to $-\infty$, which makes it problematic, because integrating over one fourth of the circle involves integrating along the imaginary line, which is not simpler than this. Any help appreciated.

$$\int_0^{\infty}\frac{\sin(s^{\frac{1}{2}})}{(1+s)^{2}} ds$$

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    $\begingroup$ You may want to first substitute $s=u^2$. $\endgroup$ – Zachary Feb 1 at 18:23
  • $\begingroup$ Wow right thank you! $\endgroup$ – ryszard eggink Feb 1 at 18:25
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Integrating by parts gives $$\int_0^{\infty}\frac{\sin(\sqrt s)}{(1+s)^{2}} ds=\int_0^\infty \frac{\cos (\sqrt s)}{1+s}\frac{ds}{2\sqrt s}$$ Now substitution $s=x^2$ and the integral become $$\int_0^\infty \frac{\cos x}{1+x^2}dx$$ Since the integrand is an even function and the real part of $e^{ix}$ is $\cos x$ we have the following equality $$\int_0^\infty \frac{\cos x}{1+x^2}dx=\frac12\int_{-\infty}^\infty \frac{e^{ix}}{(x+i)(x-i)}dx$$ Choosing the contour to be the upper semi-plane where only the pole $z=i$ exists for the function $$f(z)=\frac{e^{iz}}{(z+i)(z-i)}$$ Give the integral to be $$\frac12 \cdot 2\pi i \lim_{z\to i} \cdot (z-i)\cdot \text{Res} f(z)=\pi i \lim_{z\to i} \frac{e^{iz}}{z+i} =\pi i \frac{e^{-1}}{2i}=\frac{\pi}{2e}$$

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    $\begingroup$ Thanks, i got the same results after a hint from the comment :) $\endgroup$ – ryszard eggink Feb 1 at 19:30
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$$\newcommand{\Res}{\operatorname*{Res}} \begin{align} \int_0^\infty\frac{\sin\left(s^{1/2}\right)}{(1+s)^2}\,\mathrm{d}s &=\int_0^\infty\frac{2s\sin(s)}{(1+s^2)^2}\,\mathrm{d}s\tag1\\ &=\int_{-\infty}^\infty\frac{s\sin(s)}{(1+s^2)^2}\,\mathrm{d}s\tag2\\ &=\int_{-\infty}^\infty\frac1{2i}\left(e^{is}-e^{-is}\right)\frac1{4i}\left(\frac1{(s-i)^2}-\frac1{(s+i)^2}\right)\mathrm{d}s\tag3\\ &=-\frac18\left[2\pi i\Res_{s=i}\left(\frac{e^{is}}{(s-i)^2}\right)-2\pi i\Res_{s=-i}\left(\frac{e^{-is}}{(s+i)^2}\right)\right]\tag4\\[6pt] &=-\frac18\left[-\frac{2\pi}e-\frac{2\pi}e\right]\tag5\\[9pt] &=\frac\pi{2e}\tag6 \end{align} $$ Explanation:
$(1)$: substitute $s\mapsto s^2$
$(2)$: use symmetry
$(3)$: write $\sin(x)$ as exponentials and apply partial fractions
$(4)$: use the contour $[-R,R]\cup Re^{i[0,\pi]}$ for $e^{is}$ which contains the singularity at $s=i$
$\phantom{(4)\text{:}}$ use the contour $[-R,R]\cup Re^{i[0,-\pi]}$ for $e^{-is}$ which contains the singularity at $s=-i$
$\phantom{(4)\text{:}}$ note that the second contour is clockwise
$(5)$: evaluate the residues
$(6)$: simplify

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