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I am interested in the following integral

$$\int_{-\infty}^{\infty}\mathrm{d}x\left(1-\exp\left[\frac{-b}{\sqrt{2\pi}a}\exp\left(-\frac{x^2}{2a^2}\right)\right]\right).$$

Does anyone know how to solve this integral?

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    $\begingroup$ Interesting. But I doubt that the analytical form exists. $\endgroup$
    – konstant
    Apr 15, 2016 at 7:13

1 Answer 1

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Using the identity $1-e^{-z}=\sum_{n\geq 1}\left(-1\right)^{n-1}\frac{z^n}{n!}$, the desired integral is $\sum_{n\geq 1}\frac{I_n}{n!}$ with $$I_n:=\int_{-\infty}^{\infty}\text{d}x\exp\left[\frac{b^n}{a^n\sqrt{2\pi}^n}\exp\left(-\frac{nx^2}{2a^2}\right)\right]=\frac{b^n}{a^n\sqrt{2\pi}^n}\sqrt{\frac{2\pi a^2}{n}}.$$Define $C:=\sqrt{2\pi a^2},\,q:=\frac{b}{a\sqrt{2\pi}}$, so the desired integral is $$C\sum_{n\geq 1}\left(-1\right)^{n-1}\frac{q^n}{n!\sqrt{n}}.$$I'm not aware of any closed-form expression for this, but it clearly converges.

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  • $\begingroup$ No hope for closed-form: $\sum _{n=1}^{\infty } \frac{(-1)^{n-1} q^n}{n! \sqrt{n}}=\mathcal{L}_s^{-1}\left[-\frac{\text{Li}_{\frac{1}{2}}\left(-\frac{q}{s}\right)}{s}\right](1)=\int_0^{2 \pi } -\frac{e^{e^{i x}} \text{Li}_{\frac{1}{2}}\left(-e^{-i x} q\right)}{2 \pi } \, dx=\int_0^{\infty } \left(\frac{1}{\sqrt{\pi } \sqrt{t}}-\frac{e^{-e^{-t} q}}{\sqrt{\pi } \sqrt{t}}\right) \, dt$ :( $\endgroup$ Sep 15, 2018 at 9:32

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