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.
2026-05-17 06:48:35.1779000515
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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}
Can't solve Improper Integral $\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{\root{x}\sin\pars{x} \over 1 + x^{2}}\,\dd x}:\ {\Large ?}}$.
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.
Hint
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.