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Question Number 133490 by mnjuly1970 last updated on 22/Feb/21

$$....advncedcalculus....$$$$prove:\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{sin\left(tan\left(x\right)\right)}{x}dx=\frac{\pi }{2}(1-\frac{1}{e})$$$$provethat:\mathbf{\Phi }={\int }_0^{\mathrm{\infty }}\frac{1-e^{-x^2}}{x^2}dx=\sqrt{\pi }$$$$$$$$$$

Answered by Dwaipayan Shikari last updated on 22/Feb/21

$$I\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{e^{-\alpha x^2}-e^{-\beta x^2}}{x^2}dx$$$$I^{\prime }\left(\alpha \right)=-{\int }_0^{\mathrm{\infty }}{e}^{-\alpha x^2}dx=-\sqrt{\frac{\pi }{4\alpha }}\Rightarrow I\left(\alpha \right)=-\sqrt{\pi \alpha }+C$$$$I\left(\beta \right)=-\sqrt{\pi \beta }+C=0\Rightarrow C=\sqrt{\pi \beta }$$$$I\left(\alpha \right)=\sqrt{\pi }(\sqrt{\beta }-\sqrt{\alpha })$$$$\alpha =0\beta =1$$$$I\left(0\right)={\int }_0^{\mathrm{\infty }}\frac{1-e^{-x^2}}{x^2}dx=\sqrt{\pi }$$

Commented by mnjuly1970 last updated on 22/Feb/21

$$grateful..mrpayan$$$$verynice..$$

Answered by mathmax by abdo last updated on 22/Feb/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{1-{\mathrm{e}}^{-{\mathrm{x}}^2}}{{\mathrm{x}}^2}\mathrm{dx}\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{1-{\mathrm{e}}^{-{\mathrm{ax}}^2}}{{\mathrm{x}}^2}\mathrm{with}\mathrm{a}>0$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ax}}^2}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\left(\sqrt{\mathrm{a}}\mathrm{x}\right)}^2}\mathrm{dx}{=}_{\sqrt{\mathrm{a}}\mathrm{x}=\mathrm{y}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\frac{\mathrm{dy}}{\sqrt{\mathrm{a}}}$$$$=\frac{1}{\sqrt{\mathrm{a}}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\mathrm{dy}=\frac{1}{\sqrt{\mathrm{a}}}\frac{\sqrt{\pi }}{2}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\mathrm{k}+\sqrt{\pi \mathrm{a}}$$$${\lim }_{\mathrm{a}\to \mathrm{o}}\mathrm{f}\left(\mathrm{a}\right)=0\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\sqrt{\pi \mathrm{a}}\mathrm{and}\mathrm{\Phi }=\mathrm{f}\left(1\right)=\sqrt{\pi }$$

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