How to evaluate this integral?
$\int_{-1}^{1} x^2 \exp(\frac{1}{x^2-1}) dx$
If possible, I want any solution in elementary/non-elementary functions.
Many thanks.
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2$\begingroup$ Here is a similar integral, maybe you can use the result there? $\endgroup$mickep– mickep2015-11-06 05:22:28 +00:00Commented Nov 6, 2015 at 5:22
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$\begingroup$ Beside numerical integration, I really do not see any closed form for this one. $\endgroup$Claude Leibovici– Claude Leibovici2015-11-06 06:31:26 +00:00Commented Nov 6, 2015 at 6:31
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$\begingroup$ In terms of the Meijer G-function, we have $I=\dfrac{\sqrt\pi}2G_{1,2}^{2,0}\bigg(1;\displaystyle{5/2\choose0,1}\bigg).$ $\endgroup$Lucian– Lucian2015-11-06 15:42:12 +00:00Commented Nov 6, 2015 at 15:42
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1 Answer
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By substituting $\cosh^2t=\dfrac1{1-x^2}$ and exploiting the parity of the integrand, we finally
arrive at $I=2\Big[I_2(1)-I_4(1)\Big]$, where $I_k(a)=\displaystyle\int_0^\infty\frac{e^{-a\cosh^2x}}{\cosh^kx}~dx.~$ Also, we notice that
$-I'_2(1)=I''_4(1)=\displaystyle\int_0^\infty e^{-a\cosh^2x}~dx,~$ where the differentiation happens with regard to
the parameter a. Using the fact that $\cosh^2x=\dfrac{1+\cosh2x}2$ along with our knowledge
of Bessel functions, we evaluate the last integral to equal $\dfrac12~K_0\bigg(\dfrac a2\bigg)~\exp\bigg(\!\!-\dfrac a2\bigg).$
Unfortunately, it would seem that the latter's first and second order anti-derivatives cannot
be expressed in terms of Bessel functions. $($I might be wrong, and I hope I am, but so far
I've had no luck in finding such a closed form for them$)$. However, not all is lost, since we
can always count on the Meijer G-function to come to our rescue. Thus, $$I=\dfrac{\sqrt\pi}2~G_{1,2}^{2,0}\bigg(1;\displaystyle{5/2\choose0,1}\bigg).$$ If someone could rewrite this last expression in terms of Bessel or Struve functions, I would
be very, very grateful.
