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Question Number 135309 by mnjuly1970 last updated on 12/Mar/21

$$....\mathcal{N}ice\mathcal{C}alculus....$$$$provethat:::$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{sin\left(ksin\alpha \right)x}{\sqrt{x}}dx=\sqrt{\frac{\pi }{k}}sin\left(\frac{\alpha }{2}\right)...$$$$$$$$$$

Answered by mathmax by abdo last updated on 12/Mar/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{ksin}\alpha \right)\mathrm{x}}{\sqrt{\mathrm{x}}}\mathrm{we}\mathrm{put}\mathrm{ksin}\alpha =\lambda \Rightarrow \mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\sin \left(\lambda \mathrm{x}\right)}{\sqrt{\mathrm{x}}}\mathrm{dx}$$$$=-\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{i}\lambda \mathrm{x}}}{\sqrt{\mathrm{x}}}\mathrm{dx}\right)\mathrm{we}\mathrm{have}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{i}\lambda \mathrm{x}}}{\sqrt{\mathrm{x}}}\mathrm{dx}{=}_{\sqrt{\mathrm{x}}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{i}\lambda {\mathrm{t}}^2}}{\mathrm{t}}\left(2\mathrm{t}\right)\mathrm{dt}$$$$=2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{i}\lambda {\mathrm{t}}^2}\mathrm{dt}=2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\left(\sqrt{\lambda \mathrm{i}}\mathrm{t}\right)}^2}\mathrm{dt}{=}_{\sqrt{\lambda \mathrm{i}}\mathrm{t}=\mathrm{z}}2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}\frac{\mathrm{dz}}{\sqrt{\lambda \mathrm{i}}}$$$$=\frac{1}{\sqrt{\mathrm{i}}\sqrt{\lambda }}\times 2.\frac{\sqrt{\pi }}{2}=\sqrt{\frac{\pi }{\lambda }}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}=\sqrt{\frac{\pi }{\lambda }}\{\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}}\}\Rightarrow$$$$\mathrm{\Phi }=\frac{1}{\sqrt{2}}\sqrt{\frac{\pi }{\lambda }}=\frac{1}{\sqrt{2}}\sqrt{\frac{\pi }{\mathrm{ksin}\alpha }}=\sqrt{\frac{\pi }{2\mathrm{ksin}\alpha }}$$$$$$

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