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Prove that $$\int_0^{\pi/2} e^{-\tan^2(x)}\,dx=\frac{\pi e}{2}\big(1-\operatorname{erf}(1)\big)$$

My Attempt
Let $u=\tan(x)\implies du=\sec^2(x)\,dx=1+u^2\,dx$ $$I=\int_0^{\pi/2} e^{-\tan^2(x)}\,dx=\int_0^\infty \frac{e^{-u^2}}{1+u^2}\,du$$ How do you proceed from here? I was thinking of integrating by parts but it doesn't seem to work. How do you solve this integral? Thank you for your time

DMcMor
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hwood87
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  • I fixed what I assume was a typo: $\frac{\pi e}{2}\big(1-\operatorname{erf}(x)\big)$ should have been $\frac{\pi e}{2}\big(1-\operatorname{erf}(1)\big)$. – DMcMor Sep 08 '21 at 00:42
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    You may employ the famous trick: Let $$I(\alpha)=\int_{0}^{\infty}\frac{e^{-\alpha u^2}}{1+u^2},\mathrm{d}u$$ and show that $$I(\alpha)-I'(\alpha)=\frac{1}{2}\sqrt{\frac{\pi}{\alpha}},\qquad I(0)=\frac{\pi}{2}.$$ This differential equation can be solved by invoking any of the standard techniques (such as the integrating factor method) to yield the desired result. – Sangchul Lee Sep 08 '21 at 00:53

1 Answers1

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Note that, $$\int_{0}^{\infty } e^{-(1+t^{2})y} dy=\frac{1}{(1+t^{2})}\\ $$ $$\int_{0}^{\infty }e^{-t^{2}a}dt=\frac{^{\sqrt{\pi}}}{2\sqrt{a}}$$ $$\int_{0}^{z}e^{-t^{2}}dt=\frac{^{\sqrt{\pi}}}{2}erf(z)$$ Therefore, the solution is $$ \int_{0}^{\infty }\frac{e^{-t^{2}}}{(1+t^{{2}})}dt=\int_{0}^{\infty }\int_{0}^{\infty } e^{-(1+t^{2})y}e^{-t^{2}}dtdy$$$$\int_{0}^{\infty }\int_{0}^{\infty }e^{-t^{2}(1+y)}e^{-y}dtdy=\int_{0}^{\infty }\frac{^{\sqrt{\pi}}e^{-y}}{2\sqrt{1+y}}dy $$ $$\sqrt{1+y}=x$$$\sqrt{\pi}\int_{1}^{\infty }e^{-x^{2}+1}dx=\frac{\pi}{2}e(1-\operatorname{erf}(1))$

Sahan Manodya
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SSS
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