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I would like to calculate: $$ \int_0^1 \sqrt{1+ a x^2} e^{-a x^2/c} \, \mathrm{d}x $$ for any positive constants $a,c>0$. I am happy with any closed expression involving any kind of special functions.

Failed attempts so far:

  • standard changes of variables seem not to effectively simplify this integral. Indeed one can change variables (e.g. $ax^2/c\to y$) to simplify the dependence on parameters, but I am not able to reduce the indefinite integration to a known one.

  • Mathematica fails

  • I am not able to find this integral (or a similar one, or those integrals that I get by change of variables) on the standard references, e.g. Gradshteyn–Ryzhik.

  • Contour integration: this I did not try extensively by lack of a reasonable contour. Anyway, the integrand has singularities $\pm \tfrac{i}{\sqrt{a}}$ (choose the standard branch cut of $\sqrt{.}$ on the negative real semi-axis) and $\infty$.

  • Exploiting the symmetries of the integrand also does not seem to be of use.

AlephBeth
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    The integral is "worse" than that of the Gaussian and does not seem to be a common expression, so don't hope too much for a ready-made function. –  Nov 03 '20 at 17:55
  • Did you try integration by parts? – terrace Nov 03 '20 at 18:03
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    The obnoxious thing about integration is how easy it is to create "simple-looking" integrals that have no closed form. This seems likely to be one of them. – eyeballfrog Nov 03 '20 at 18:07
  • @terrace let me confess straight away that Mathematica was my first attempt. I seem to recall that IbP is a standard routine in Mathematica integration, so I did not try it myself. I tried standard ODE methods in the parameters though (e.g. differentiate w.r.t. a, see if the quantity solves and ODE in a and try to solve that), but this was also no use – AlephBeth Nov 03 '20 at 18:08
  • @YvesDaoust indeed this does not look too nice. I was hoping to have some expression in terms of hypergeometric functions. For instance the same integral from $0$ to $\infty$ evaluates to a precise hypergeometric function of the parameters... – AlephBeth Nov 03 '20 at 18:10
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    Even for $a=c=1$, the value of the definite integral $$\displaystyle\int_0^1 \sqrt{1+x^2}e^{-x^2} , \mathrm{d}x ,, \approx 0.8319314795 \dots$$ does not show up on the Inverse Symbolic Calculator. – Zubin Mukerjee Nov 03 '20 at 18:11

1 Answers1

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$\int_0^1\sqrt{1+ax^2}~e^{-\frac{ax^2}{c}}~dx$

$=\int_0^1\sqrt{1+ax}~e^{-\frac{ax}{c}}~d(\sqrt{x})$

$=\dfrac{1}{2}\int_0^1x^{-\frac{1}{2}}(1+ax)^\frac{1}{2}e^{-\frac{ax}{c}}~dx$

$=\Phi_{1}\left(\dfrac{1}{2},-\dfrac{1}{2},\dfrac{3}{2};-a,-\dfrac{a}{c}\right)$ (according to https://en.wikipedia.org/wiki/Humbert_series)

Harry Peter
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  • Brilliant! Thank you very much! I am also impressed by the simplification of all constants in front of the integral... – AlephBeth Nov 07 '20 at 19:46