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Question Number 150828 by mnjuly1970 last updated on 15/Aug/21

$$$$$${\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(x\right)}{x\sqrt{x}}dx\stackrel{?}{=}\sqrt{\pi }$$

Answered by puissant last updated on 16/Aug/21

$$Q={\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}x}{x\sqrt{x}}dx$$$$={\int }_0^{\mathrm{\infty }}{x}^{-\frac{3}{2}}{sin}^{2}xdx$$$$={\left[\frac{1}{1-\frac{3}{2}}{x}^{-\frac{1}{2}}{sin}^{2}x\right]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}-2x^{-\frac{1}{2}}\times 2sinxcosxdx$$$$=2{\int }_0^{\mathrm{\infty }}\frac{sin2x}{\sqrt{x}}dx$$$$\underset{\sqrt{x}\to u}{=}2{\int }_0^{\mathrm{\infty }}\frac{sin2u^2}{u}\left(2udu\right)$$$$=4{\int }_0^{\mathrm{\infty }}sin2u^{2}du=-4im\left({\int }_0^{\mathrm{\infty }}{e}^{-2{iu}^2}du\right)$$$$\because \because \because {\int }_0^{\mathrm{\infty }}{e}^{-{\left(\sqrt{2i}u\right)}^2}du\underset{\sqrt{2i}u\to z}{=}{\int }_0^z{e}^{-z^2}\frac{dz}{\sqrt{2i}}$$$$=\frac{1}{\sqrt{2}}{e}^{-i\frac{\pi }{4}}\times \frac{\sqrt{\pi }}{2}$$$$=\frac{\sqrt{\pi }}{2\sqrt{2}}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=\frac{\sqrt{\pi }}{4}-i\frac{\sqrt{\pi }}{4}$$$$Q=-4im\left({\int }_0^{\mathrm{\infty }}{e}^{-2{iu}^2}du\right)=-4(-\frac{\sqrt{\pi }}{4})$$$$\because \therefore Q=\sqrt{\pi }..$$$$$$$$............\mathcal{L}epuissant..........$$

Commented by mnjuly1970 last updated on 16/Aug/21

$$thankyousomuch...$$

Answered by mathmax by abdo last updated on 17/Aug/21

$$\mathrm{{\rm Y}}={\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\left(\mathrm{x}\right)}{\mathrm{x}\sqrt{\mathrm{x}}}\mathrm{dx}{=}_{\sqrt{\mathrm{x}}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\left({\mathrm{t}}^2\right)}{{\mathrm{t}}^3}\left(2\mathrm{t}\right)\mathrm{dt}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\left({\mathrm{t}}^2\right)}{{\mathrm{t}}^2}\mathrm{dt}=2\{{[-\frac{1}{\mathrm{t}}{\sin }^2\left({\mathrm{t}}^2\right)]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}-\frac{1}{\mathrm{t}}\times 2\sin \left({\mathrm{t}}^2\right)2\mathrm{t})\cos \left({\mathrm{t}}^2\right)\mathrm{dt}\}$$$$=2{\int }_0^{\mathrm{\infty }}\sin \left({2\mathrm{t}}^2\right)\mathrm{dt}=2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\sin \left({2\mathrm{t}}^2\right)\mathrm{dt}=-2\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-{2\mathrm{it}}^2}\mathrm{dt}\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-\left(\sqrt{2}\mathrm{i}\right){\mathrm{t}}^2}\mathrm{dt}{=}_{\sqrt{2}\mathrm{it}=\mathrm{y}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\frac{\mathrm{dy}}{\sqrt{2}\mathrm{i}}$$$$=\frac{\sqrt{\pi }}{\sqrt{2}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}}=\frac{\sqrt{\pi }}{\sqrt{2}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}=\frac{\sqrt{\pi }}{\sqrt{2}}(\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}})=\frac{\sqrt{\pi }}{2}-\frac{\mathrm{i}\sqrt{\pi }}{2}\Rightarrow$$$$\mathrm{{\rm Y}}=2\times \frac{\sqrt{\pi }}{2}=\sqrt{\pi }$$

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