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Question Number 190987 by Rupesh123 last updated on 15/Apr/23

Answered by 07049753053 last updated on 16/Apr/23

$$\mathbf{let}{\mathbf{x}}^2=\mathbf{u}\mathbf{dx}=\frac{\mathbf{du}}{2\sqrt{\mathbf{u}}}$$$$\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathbf{e}}^{-\mathbf{u}}\mathbf{sin}\left(\mathbf{u}\right)}{\mathbf{u}\sqrt{\mathbf{u}}}\mathbf{du}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{\mathbf{u}}^{-\frac{3}{2}}{\mathbf{e}}^{-\mathbf{u}}\mathbf{sin}\left(\mathbf{u}\right)\mathbf{du}$$$$\mathbf{by}{\mathbf{euler}}^{\prime }\mathbf{s}\mathbf{formula}$$$$\mathbf{sin}\left(\mathbf{u}\right)=\mathcal{I}\mathit{m}\left({\mathbf{e}}^{\mathbf{iu}}\right)$$$$\frac{1}{2}\mathcal{I}\mathit{m}{\int }_0^{\mathrm{\infty }}{\mathbf{u}}^{-\frac{3}{2}}{\mathbf{e}}^{-\mathbf{u}(\mathbf{i}-1)}\mathbf{du}$$$$\mathbf{let}\mathbf{u}(\mathbf{i}-1)=\mathbf{k}\mathbf{u}=\frac{\mathbf{k}}{\mathbf{i}-1}\mathbf{du}=\frac{\mathbf{dk}}{\mathbf{i}-1}$$$$\frac{1}{2}\mathcal{I}\mathit{m}\left(\frac{1}{{(\mathbf{i}-1)}^{-\frac{1}{2}}}{\int }_0^{\mathrm{\infty }}{\mathbf{k}}^{-\frac{3}{2}}{\mathbf{e}}^{-\mathbf{k}}\mathbf{dk}\right)$$$$\frac{1}{2}\mathcal{I}\mathbf{m}\left(\frac{1}{{(\mathbf{i}-1)}^{-\frac{1}{2}}}\mathbf{\Gamma }(-\frac{1}{2})\right)=\frac{1}{2}\mathcal{I}\mathbf{m}\left(\sqrt{(1-i)}\right)\mathbf{\Gamma }(-\frac{1}{2})$$$$\frac{1}{2}\left[\sqrt{\mathit{\pi }}\sqrt{\frac{1}{2}-\frac{1}{\sqrt{2}}}\right]=\frac{\sqrt{\mathit{\pi }}}{2}\sqrt{\frac{\sqrt{2}-2}{2\sqrt{2}}}\approx 0.806625...$$$$\mathcal{S}\mathit{m}\mathit{a}\mathit{l}\mathit{l}\mathit{L}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}$$

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