I am trying to calculate the following integral:

$$\int_{-1}^{1}\ln(1+\gamma x)\exp\left(-\frac{(z-x)^2}{2}\right)dx$$

with $\gamma\in[-1,1]$, $z\in\mathbb{R}$ and ln being the natural logarithm. I tried a lot of thinks like substitution, integration by parts, used the series expansion of the natural logarithm resp of the exponential function. I also searched for some help in Literatur but couldn't find anything helpful. My question is: Can we solve this integral?

  • $\begingroup$ Even for $\gamma=0$ you have a sum of $\textrm{erf}$ functions. So I doubt that you will have a nice analytical result. You can solve it numerically $\endgroup$ – Andrei Dec 28 '20 at 15:33
  • $\begingroup$ @Andrei , for $\gamma = 0$ the integrand is 0. gimpfel, are you looking for a closed form answer? or is an infinite series representation enough? $\endgroup$ – Tom Ariel Dec 28 '20 at 15:41
  • 3
    $\begingroup$ @TomAriel Oops. Sorry, my mistake.$\ln(1)\ne 1$, not even after Christmas :) $\endgroup$ – Andrei Dec 28 '20 at 15:43
  • $\begingroup$ @TomAriel do you have an idea about an infinite series representation ? $\endgroup$ – gimpfel Dec 29 '20 at 20:20
  • $\begingroup$ @gimpfel If an answer was helpful to you, please don't forget to 'accept' it, so this question is marked as resolved. $\endgroup$ – Sal Feb 1 at 14:04

Considering first the restricted problem with $z=0$.

We can find an approximate series for $\gamma\rightarrow 0$. First, expand the logarithm

$$\ln(1+\gamma x)=\gamma x - \frac{(\gamma x)^2}{2}+ ... $$

Now notice that on the integration domain, all terms $x^{odd}$ will integrate to zero. We are left with

$$ I(\gamma)=-\int_{-1}^{1}dx\ \left( \frac{(\gamma x)^2 }{2}+ \frac{(\gamma x)^4 }{4}+... \right)e^{-x^2/2}$$

Term by term integration gives

$$ I(\gamma)\sim\sum_{n=1}^\infty A_n \gamma^{2n}, \ \gamma\rightarrow 0$$

With the coefficients

$$ A_n =-\frac{2^{n-1/2}}{n}\left( \Gamma(n+1/2)-\Gamma(n+1/2,1/2) \right)$$

The degree to which this is an improvement depends on how you feel about the incomplete gamma function. By repeated use of the identities

$$\Gamma(s+1,x)=s \Gamma(s,x)+x^s e^{-x} $$

$$ \Gamma(1/2,x)=\sqrt{\pi} \operatorname{erfc}(\sqrt{x})$$

Every term in the series may be cast into the form of $\operatorname{elementary function} \times \operatorname{erfc}(2^{-1/2}) $. This is an improvement because now all those $\operatorname{erfc}$s appear as constants in the series. The first term alone is a fine approximation

$$ I(\gamma)\sim -\gamma^2 \left(\sqrt{\pi/2} \left( 1-\operatorname{erfc}(2^{-1/2})\right)-e^{-1/2} \right),\gamma\rightarrow0$$

More importantly we learn that for small $\gamma$, the integral scales as $-\gamma^2$.

Here is a plot of the first term together with the exact (numerically integrated) result

enter image description here

And the same plot with just three terms

enter image description here

We can also (somewhat) extend this to the case $z\rightarrow 0$.

To avoid issues of non-uniform convergence, I'll consider the integral

$$I(\epsilon)= \int_{-1}^1 \ dx \ln(1+\gamma \epsilon x) \exp \left(-\frac{(x-\epsilon z)^2}{2}\right) $$

This of course is still restricted compared to the original. Expanding the exponential around $\epsilon=0$ will produce a series of 'corrections' to the $\ln$ series. The first nonzero term still occurs at $\epsilon^2$. I find

$$ I(\epsilon)\sim -\epsilon^2 \frac{\gamma(\gamma-2z)}{2\sqrt{e}}\left( \sqrt{2e\pi} \operatorname{erf}(2^{-1/2})-2\right),\epsilon\rightarrow 0$$

We learn the integral scales with $-\gamma(\gamma-2z)$ for small $\epsilon$. Here's a plot with $\gamma=1/2$

enter image description here

As expected, when $z>1$ the approximation is bad: this is when the peak of the Gaussian passes outside of the integral bounds.

For large $z$, I think we should be able to write the exponential in such a way as to use Laplace's method, but I don't see how right now.


$$I=\int\log(1+\gamma x)\,\exp\left(-\frac{(z-x)^2}{2}\right)dx$$

Let $x=\sqrt{2} t+z$ to make $$I=\sqrt{2}\int \log \big[1+\gamma \left(\sqrt{2} t+z\right)\big]\,e^{-t^2}\,dt$$ Using Taylor around $t=0$ $$ \log \big[1+\gamma \left(\sqrt{2} t+z\right)\big]=\log (1+\gamma z)-\sum_{n=1}^\infty \frac{2^{n/2} \left(-\frac{\gamma }{\gamma z+1}\right)^n}{n} t^n$$ and $$J_n=\int t^n\,e^{-t^2}\,dt=-\frac{1}{2} \Gamma \left(\frac{n+1}{2},t^2\right)$$

This would be a nightmare.


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