# Can't solve Improper Integral $\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$

Whilst checking for the existence of improper integrals, I came across this one: $$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$ So in order to check its existence I simply have to see if the limit: $$\lim_{a\to\infty} \int_{0}^{a} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$ Is a number or not.
However I seem unable to find a way to solve this particular Integral, and neither any online calculator can. I have tried all substitutions that I could think of, as well as partial integration and using any helpful trigonometric identities but they were all in vain.

• Do you simply want to know if the integral converges, or what it converges to? – user455343 Aug 11 '17 at 21:17
• I have tried integration by parts with $(arctan(x))'=\frac{1}{1+x^2}$ but it didn't work, any other which you would suggest? – Konstantinos Zafeiris Aug 11 '17 at 21:18
• @MatthewStonebraker Only knowing whether it converges or not would be fine – Konstantinos Zafeiris Aug 11 '17 at 21:19
• Not sure on accuracy or helpfulness, but Mathematica gives $$\int_0^\infty\frac{\sqrt{x}\sin(x)}{1+x^2}\,dx=\frac{\pi\left(-e^2 \text{erfc}(1) + \text{erfi}(1)+1\right)}{2\sqrt 2 e} \approx 0.608068$$ – Dando18 Aug 11 '17 at 21:24
• Convergence is easy, because the integrand is bounded in absolute value by $x^{-3/2}$ for large $x$. Use comparison. – user296602 Aug 11 '17 at 21:24

If $f$ is continuous in $I=(1, +\infty)$ and if $x^{1+\epsilon} f(x)$ is bounded in $I$ for some $\epsilon > 0$, then $\int_1^\infty f(x) dx$ converges.
The integration is "closed" along a $$\ds{45^{\large\circ}}$$-pizza slice contour in the complex plane first quadrant. Namely, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty} {\root{x}\sin\pars{x} \over 1 + x^{2}}\,\dd x} \,\,\,\stackrel{x\ \mapsto\ x^{2}}{=}\,\,\, 2\int_{0}^{\infty} {x^{2}\sin\pars{x^{2}} \over 1 + x^{4}}\,\dd x \\[5mm] = &\ 2\,\Im\int_{0}^{\infty}{x^{2}\expo{\ic x^{2}} \over 1 + x^{4}}\,\dd x \\[5mm] = & \lim_{\epsilon \to 0^{\large +}}\left[-2\,\Im\int_{\infty}^{1 + \epsilon}{r^{2}\,\ic\,\exp\pars{-r^{2}} \over 1 - r^{4}}\,\expo{\ic\pi/4}\,\dd r\right. \\[2mm] &\ \phantom{\lim_{\epsilon \to 0^{\large +}}\left[\right.} - 2\,\Im\int_{\pi/4}^{-3\pi/4}{\ic\expo{-1} \over 1 - \pars{\expo{\ic\pi/4} + \epsilon\expo{\ic\theta}}^{4}}\,\epsilon\expo{\ic\theta}\ic \,\dd\theta \\[2mm] & \phantom{\lim_{\epsilon \to 0^{\large +}}\left[\right.} \left. -2\,\Im\int_{1 - \epsilon}^{0}{r^{2}\,\ic\,\exp\pars{-r^{2}} \over 1 - r^{4}}\,\expo{\ic\pi/4}\,\dd r\right] \\[5mm] = &\ 2\mrm{P.V.}\Im\int_{0}^{\infty}{r^{2}\,\ic\,\exp\pars{-r^{2}} \over 1 - r^{4}}\,\expo{\ic\pi/4}\,\dd r \\[2mm] &\ + 2\lim_{\epsilon \to 0^{+}}\Im\int_{-3\pi/4}^{\pi/4} {\ic\,\expo{-1} \over -4\expo{-\ic\pi/4}\epsilon\expo{\ic\theta}}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta \\[5mm] = &\ \root{2}\mrm{P.V.}\int_{0}^{\infty} {r^{2}\exp\pars{-r^{2}} \over 1 - r^{4}}\,\dd r + {1 \over 4}\,\root{2}\pi\expo{-1} \\[5mm] = &\ {\root{2} \over 2}\,\ \underbrace{\mrm{P.V.}\int_{0}^{\infty} {\exp\pars{-r^{2}} \over 1 - r^{2}}\,\dd r} _{\ds{\pi\,\mrm{erfi}\pars{1} \over 2\expo{}}} - {\root{2} \over 2}\ \underbrace{\int_{0}^{\infty}{\exp\pars{-r^{2}} \over 1 + r^{2}}\,\dd r}_{\ds{{\pi \over 2}\expo{}\,\mrm{erfc}\pars{1}}}\ \\ &\ + {1 \over 4}\,\root{2}\pi\expo{-1} \\[5mm] = &\ \bbx{{\root{2} \over 4}\,\pi\expo{-1}\bracks{\mrm{erfi}\pars{1} + 1} -{\root{2} \over 4}\,\pi\expo{}\,\mrm{erfc}\pars{1}}\ \approx\ 0.6081 \\ & \end{align}
$$\ds{\mrm{erfi}\ \mbox{and}\ \mrm{erfc}}$$ are Error Functions.