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Question Number 154409 by mnjuly1970 last updated on 18/Sep/21

$$$$$$nicecalculus..$$$$provethat:$$$$$$$$\mathrm{I}:={\int }_0^{\mathrm{\infty }}\frac{(1+e^{-x}){sin}^2\left(x\right)}{x^{\frac{3}{2}}}=\sqrt{2\pi }(1+\sqrt{\sqrt{2}-1})$$$$m.n$$

Answered by Kamel last updated on 18/Sep/21

$$$$$$\mathrm{I}:={\int }_0^{\mathrm{\infty }}\frac{(1+e^{-x}){sin}^2\left(x\right)}{x^{\frac{3}{2}}}$$$$=2{\int }_0^1{\int }_0^{+\mathrm{\infty }}{x}^{-\frac{1}{2}}(1+e^{-x})sin\left(2ax\right)dxda$$$$=2{\int }_0^1(\sqrt{2}{a}^{-\frac{1}{2}}{\int }_0^{+\mathrm{\infty }}sin\left(u^2\right)du+\frac{1}{2i}{\int }_0^{+\mathrm{\infty }}(e^{-(1-2ai)x^2}-e^{-(1+2ai)x^2})dx)da$$$$=2{\int }_0^1(2^{-\frac{1}{2}}{a}^{-\frac{1}{2}}\frac{\sqrt{\pi }}{2\sqrt{2}}+\frac{\sqrt{\pi }}{4i}(\frac{1}{\sqrt{1-2ai}}-\frac{1}{\sqrt{1+2ai}}))da$$$$=\sqrt{\pi }\left(\frac{1}{2}(\sqrt{1-2i}+\sqrt{1+2i})\right)=\sqrt{2\pi (1+\sqrt{5})}$$

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