Calculate $\int_{0}^{\infty }e^{-x^{2}-x^{-2}}dx$

Question

How to calculate $$\int_{0}^{\infty }e^{-x^{2}-x^{-2}}dx$$ I have no idea where to start.Is it connect with the euler-poisson integral?

Answer 1

In fact,let see the more general form $$I=\int_{0}^{\infty }e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}x$$ let $x\rightarrow x^{-1}$, we have $$I=\int_{0}^{\infty }e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}x=\int_{0}^{\infty }x^{-2}e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}x$$ hence \begin{align*} I=\frac{1}{2}\int_{0}^{\infty }\left ( 1+x^{-2} \right )e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}x&=\frac{1}{2}\int_{-\infty }^{\infty }e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}\left ( x-x^{-1} \right ) \\ &=\frac{1}{2}\int_{-\infty }^{\infty }e^{-\alpha ^{2}\left [ \left ( x-x^{-1} \right )^{2}-2 \right ]}\,\mathrm{d}\left ( x-x^{-1} \right ) \\ &=\frac{1}{2}e^{-2\alpha ^{2}}\int_{-\infty }^{\infty }e^{-\alpha ^{2}t^{2}}\,\mathrm{d}t \\ &=e^{-2\alpha ^{2}}\cdot \frac{\sqrt{\pi }}{2\alpha } \end{align*} now let $\alpha =1$ and the answer will follow.