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I recently got stuck on evaluating the following integral, $$\int_{-\infty}^{\infty} e^{-x^2/a}\ln\left(1 + be^{-cx^2}\right)dx$$ where $a>0$, $b>0$. I don't know if there is an effective substitution to use.

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  • $\begingroup$ Are you sure you don't mean $a,c>0$ instead of $a,b>0$? $\endgroup$ – David H Nov 21 '13 at 5:30
  • $\begingroup$ nah, $c$ can be a negative number. $\endgroup$ – aniki_aishiteru Nov 21 '13 at 5:31
  • $\begingroup$ well, I really should put $-cx^2$ instead of $-x^2/c$ $\endgroup$ – aniki_aishiteru Nov 21 '13 at 5:38
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First assume that $c > 0$. Simply plugging the series expansion

$$\log(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{n}, \quad |x| < 1$$

and integrating term by term, we have

$$ \int_{-\infty}^{\infty} e^{-x^{2}/a} \log ( 1 + b e^{-c x^{2}} ) \, dx = \sqrt{a \pi} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n\sqrt{acn + 1}} b^{n}, $$

which is valid for $a, c > 0$ and $|b| < 1$. Of course, the left-hand side extends holomorphically for any complex $b$ avoiding the cut $(-\infty, -1)$.

For $c < 0$, the behavior changes completely. For $b > 1$, we write

$$ \int_{-\infty}^{\infty} e^{-x^{2}/a} \log ( 1 + b e^{-cx^{2}} ) \, dx = \int_{-\infty}^{\infty} e^{-x^{2}/a} \left\{ |c|x^{2} + \log b + \log ( b^{-1} e^{-|c|x^{2}} + 1 ) \right\} \, dx. $$

The applying a similar technique,

$$ \int_{-\infty}^{\infty} e^{-x^{2}/a} \log ( 1 + b e^{-cx^{2}} ) \, dx = \sqrt{a\pi} \left( \frac{a|c|}{2} + \log b + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \sqrt{a|c|n + 1}} b^{-n} \right). $$

Edit. Actually we may assume $a = 1$ upon applying the substitution $x \mapsto \sqrt{a} x$ and absorbing $\sqrt{a}$ to $c$. Then we have the following formal series expansion with respect to $c$:

$$ \int_{-\infty}^{\infty} e^{-x^{2}} \log ( 1 + b e^{-cx^{2}} ) \, dx = -\pi \sum_{k=0}^{\infty} \frac{\operatorname{Li}_{1-k}(-b)}{k!\Gamma\left(\frac{1}{2}-k\right)} c^{k} \quad \text{as } c \to 0. $$

This series asymptotic in the sense that it diverges for any $c \neq 0$, but for any $n$ its first finite $n$ terms gives an approximation of the integral with an error $O(c^{n+1})$.

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  • $\begingroup$ Thanks for the answer. I am still looking for a closed form solution from the fat Russian book. Maybe some substitutions can do the job. If this is not solvable then I guess that I have to use the series expansion. :( $\endgroup$ – aniki_aishiteru Nov 21 '13 at 6:03
  • $\begingroup$ thanks for the answer. Do you think there will be a closed form solution for this? $\endgroup$ – aniki_aishiteru Nov 21 '13 at 10:01
  • $\begingroup$ @aniki_aishiteru, Actually I'm skeptical about the existence of a nice, simple closed form. $\endgroup$ – Sangchul Lee Nov 21 '13 at 10:14

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