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This question already has been solved here: How to evaluate $\int_{0}^{\infty}\exp(-x^2-1/x^2)dx$?

And there is also an answer for How should I calculate $\int_0^\infty e^{-\frac{1}{2}(x^2+a^2/x^2)}\,dx$ But I think the answer is incorrect, if this problem solved, and I can try to replace x by u^2 in my question, even this would still be a little different because there would have another more 2u in the equation, but would it still be some help to solve my problem?

Martin Zhang
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    Please add more context. You should mention in your body of question the integral which you wish to be evaluated (I am guessing it's the one in title). Also do you think the linked question will help here? Do you want a solution similar to the linked question? Please also share your own ideas/attempts at evaluating the integral. Questions which are just problem statements are discouraged here. – Paramanand Singh Mar 24 '21 at 05:26

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The answer can be expressed in terms of Bessel functions, thus: $$\fbox{$2 \sqrt{a} K_1\left(2 \sqrt{a}\right)\text{ if }\Re(a)>0$},$$ since one of the standard integral formulas for Bessel K functions is

$$K_a(x) = \int_0^\infty \exp(-x \cosh t) \cosh a t \, dt$$ The substitution is fairly obvious.

metamorphy
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Igor Rivin
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  • Thanks for your answer, but I am still confused, could you please spare some time for a more detailed explanation? – Martin Zhang Mar 24 '21 at 07:37
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    @MartinZhang: substitution $x=a^{1/2}e^t$, symmetrization w.r.t. $t$, reference. – metamorphy Mar 24 '21 at 09:38
  • @metamorphy: I have done according to your instruction, $$\int_{0}^{\infty}\exp(-x-a/x)dx= \int_0^\infty \exp(-2 \sqrt{a} \cosh t) \ (2 \sqrt{a})\ (exp(t)/2), dt$$, which is slightly different from $$2 \sqrt{a} K_1\left(2 \sqrt{a}\right)=\int_{0}^{\infty}\exp(-2 \sqrt{a} \cosh t) \ (2 \sqrt{a})\ ( (exp(t)+exp(-t) )/2), dt$$. Is there anything wrong in my calculation? Thanks! – Martin Zhang Mar 25 '21 at 08:47
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    @MartinZhang: After the substitution, you should have got $\int_{\color{red}{-\infty}}^\infty$. Then you split it as $\int_{-\infty}^0+\int_0^\infty$, do $t\mapsto-t$ in the first integral (that's what I meant by "symmetrization"), and arrive at the expected. – metamorphy Mar 25 '21 at 09:52
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    @metamorphy@Igor Rivin: Thanks very much for your help, I could never solve this without your help! – Martin Zhang Mar 26 '21 at 02:13