$$\int_0^\infty e^{-\frac{1}{2}(x^2+a^2/x^2)}\,dx$$
I even know the answer but indefinite integral is much more complicated than answer. The answer, according to Wolfram is $\sqrt{\pi/2}\,e^{-a}$.
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How can a definite integral w.r.t. $x$ have an $x$ in the answer? – Adrian Keister Dec 06 '19 at 20:37
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Oops, excuse me, in answer should be a instead of x. – Matthew Zhukov Dec 06 '19 at 20:38
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Let $$I=\int_0^\infty e^{-\frac{1}{2}(x^2+a^2/x^2)}\,dx. $$ Then under $x\to a^2x$, one has $$I=\int_0^\infty a^2e^{-\frac{1}{2}a^2(x^2+1/x^2)}\,dx=\int_0^\infty e^{-\frac{1}{2}a^2(x-1/x)^2}\,dx$$ Under $x-\frac1x \to x$, one can get the answer.
xpaul
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