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Does analytical solution to following integral exist?

$$\int_0^\infty { \frac{\exp\left[ {-\left(\frac{1}{t}+t+t^3\right)}\right]}{\sqrt{t}} }\mathrm{d}t$$

It converges numerically. I tested in Mathematica. So it looks like there is some kind of coordinate transformation that makes the pole at t=0 integrable. I cannot see which kind of transformation does it. I am looking for the analytical result. Can anyone help me?

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  • $\begingroup$ are all the t in the numerator within the exp function? $\endgroup$ – Cato Mar 29 '17 at 11:57
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By setting $t=x^2$ the given integral becomes $$ \int_{-\infty}^{+\infty}\exp\left[-\left(\frac{1}{x^2}+x^2+x^6\right)\right]\,dx \tag{1}$$ and by Glasser's master theorem (it is enough to set $x-\frac{1}{x}=u$) $$ \int_{-\infty}^{+\infty}\exp\left[-\left(\frac{1}{x^2}+x^2\right)\right]\,dx = \frac{\sqrt{\pi}}{2e^2}\tag{2}$$ so my suggestion is to perform the same substitution ($x-\frac{1}{x}=u$) in $(1)$ then switch to a series expansion. It looks that $(1)$ does not have a nice closed form.

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    $\begingroup$ I am amazed at the number of very fine results you know in Analysis ! $\endgroup$ – Jean Marie Mar 29 '17 at 13:27
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    $\begingroup$ @JeanMarie, legend has it that Jack is 180+ years old. $\endgroup$ – Zaid Alyafeai Mar 29 '17 at 15:55
  • $\begingroup$ @Zaid Alyafeai Good humor ! $\endgroup$ – Jean Marie Mar 29 '17 at 20:34
  • $\begingroup$ @FelixMarin: $e^{-x^6}\leq 1$. $\frac{\sqrt{\pi}}{2e^2}$ is just the first term of the alternating series giving $(1)$. $\endgroup$ – Jack D'Aurizio Mar 30 '17 at 9:03
  • $\begingroup$ Fine. I got it. $\endgroup$ – Felix Marin Mar 30 '17 at 19:52

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