Finding an expression for $\int_{-\infty}^\infty e^{-x^2/2}\ln(\cosh(ax))dx$

Question

I want to reduce the following integral into something of the form below: $$\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}} e^{-x^2/2}\ln(\cosh(a\sqrt{b}x))dx=\frac{a^2}{2}b-\frac{a^2}{4}b^2+O(b^3),$$ where $O$ is the big $O$ notation. I believe the "coefficients" for $b$ and $b^2$ are $a^2/2$ and $-a^2/4$, respectively. The exact expression for the $O(b^3)$ is not important but only that the rest of terms are $\sim O(b^3)$. I am not quite sure how to get the desired equality.

Answer 1

Here is a result of heuristic computation: If we write $\alpha = a\sqrt{b}$, then the integral has the following formal asymptotic expansion

\begin{align*} I(\alpha) &:=\int_{-\infty}^{\infty} e^{-x^2/2}\log\cosh(\alpha x) \, dx \\ &\hspace{2em} `` =\!\text{''} \ -\sqrt{2\pi} \sum_{n=1}^{\infty} \frac{B_{2n}}{n!n} 2^{n-1}(1-2^{2n}) \alpha^{2n} \\ &\hspace{2.2em} = \sqrt{2\pi} \left( \frac{1}{2} \alpha^2 - \frac{1}{4}\alpha^4 + \frac{1}{3}\alpha^6 - \frac{17}{24}\alpha^8 + \frac{31}{15}\alpha^{10} + \cdots \right) \end{align*}

This is only in a formal sense, which should be understood as

$$ I(\alpha) = -\sqrt{2\pi} \sum_{n=1}^{N} \frac{B_{2n}}{n!n} 2^{n-1}(1-2^{2n}) \alpha^{2n} + \mathcal{O}_N(\alpha^{2N+2}) $$

for each given $N \geq 1$. This is simply because the given infinite summation converges only when $\alpha = 0$.

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For a justification, it is not hard to check that $I$ is a smooth function on $\mathbb{R}$ and its derivatives can be computed by applying Leibniz's integral rule. So in principle we can compute the Taylor approximation of $I$ around $\alpha = 0$. The above result is indeed the outcome of this computation. To put simple, the above asymptotic expansion is just a consequence of the Taylor's theorem (modulo figuring out the exact coefficients).

Answer 2

This is more of a long comment, but it might be helpful.

Observe that $\cosh$ and $x^2$ are even, so let's look at $$ \int_0^\infty e^{-x^2/2}\ln(\cosh(a\sqrt{b}x))dx. $$ Since $$ \cosh(a\sqrt{b}x)=\frac{e^{a\sqrt{b}x}+e^{-a\sqrt{b}x}}{2}, $$ it follows that \begin{align} \int_0^\infty e^{-x^2/2}\ln(e^{a\sqrt{b}{x}})dx-\sqrt{\frac{\pi}{2}}\ln 2\leq \int_0^\infty e^{-x^2/2}\ln(\cosh(a\sqrt{b}x))dx\leq \int_0^\infty e^{-x^2/2}\ln(e^{a\sqrt{b}{x}})dx \end{align} Everything here can be evaluated with substitution, but I'm getting something closer to $a\sqrt{b}$ and not what you're looking for. The difference might depend on whether you are thinking of $a$ and $b$ as large or as small.

Answer 3

Too long for a comment.

Just out of curiosity, starting from Sangchul Lee's answer, I computed the integral for the orders given. $$\left( \begin{array}{ccccccc} \alpha & I_2 & I_4 & I_6 & I_8 & I_{10} & \text{exact} \\ 0.05 & 0.00125 & 0.0012484 & 0.0012484 & 0.0012484 & 0.0012484 & 0.0012484 \\ 0.10 & 0.00500 & 0.0049750 & 0.0049753 & 0.0049753 & 0.0049753 & 0.0049753 \\ 0.15 & 0.01125 & 0.0111234 & 0.0111272 & 0.0111271 & 0.0111271 & 0.0111271 \\ 0.20 & 0.02000 & 0.0196000 & 0.0196213 & 0.0196195 & 0.0196197 & 0.0196197 \\ 0.25 & 0.03125 & 0.0302734 & 0.0303548 & 0.0303440 & 0.0303460 & 0.0303456 \\ 0.30 & 0.04500 & 0.0429750 & 0.0432180 & 0.0431715 & 0.0431837 & 0.0431808 \\ 0.35 & 0.06125 & 0.0574984 & 0.0581112 & 0.0579517 & 0.0580087 & 0.0579916 \\ 0.40 & 0.08000 & 0.0736000 & 0.0749653 & 0.0745011 & 0.0747178 & 0.0746408 \\ 0.45 & 0.10125 & 0.0909984 & 0.0937664 & 0.0925753 & 0.0932790 & 0.0929918 \\ 0.50 & 0.12500 & 0.1093750 & 0.1145830 & 0.1118160 & 0.1138350 & 0.1129120 \end{array} \right)$$ It is quite clear that, if the approximation is very good for small values of parameter $\alpha$, it becomes quite poor as soon as $\alpha >0.35$.

Adding more terms does not make the situation better. Below is a table for $\alpha =0.5$ $$\left( \begin{array}{cc} n & I_n \\ 2 & 0.125000\\ 4 & 0.109375 \\ 6 & 0.114583 \\ 8 & 0.111816 \\ 10 & 0.113835 \\ 12 & 0.111960 \\ 14 & 0.114076 \\ 16 & 0.111262 \\ 18 & 0.115571 \\ 20 & 0.108105 \end{array} \right)$$

Edit

Looking at much larger values of parameter $\alpha$, i.e $(1 \leq\alpha\leq 100)$, the integral varies almost exactly as a linear function of $\alpha$ as Michael Burr answered $(R^2=1.)$

Answer 4

Again, too long for a comment.

We could get better approximation if, instead of using Taylor to expand $\log(\cosh(\alpha x))$, we use the Padé approximant $$\log(\cosh(y))=\frac{\frac{1}{2}y^2+\frac{241 }{3060}y^4-\frac{11 }{2295}y^6+\frac{883 }{1927800}y^8 } {1+\frac{248 }{765}y^2}$$ which is very good up to $y=2$.

This would give for the integral the approximation $$I=\frac{129384469125}{211833552896}+\frac{257953751 }{854167552}\alpha^2-\frac{197801 }{3444224}\alpha^4+\frac{883 }{41664}\alpha^6-\frac{388153407375}{847334211584 }\sqrt{\frac{85 \pi }{31}}\frac{e^{\frac{765}{496 \alpha^2}}}\alpha\text{erfc}\left(\frac{3 \sqrt{\frac{85}{31}}}{4 \alpha}\right)$$ For the same values as before, the results are $$\left( \begin{array}{ccc} \alpha & \text{approximation} & \text{exact} \\ 0.05 & 0.0012484 & 0.0012484 \\ 0.10 & 0.0049753 & 0.0049753 \\ 0.15 & 0.0111271 & 0.0111271 \\ 0.20 & 0.0196197 & 0.0196197 \\ 0.25 & 0.0303456 & 0.0303456 \\ 0.30 & 0.0431809 & 0.0431808 \\ 0.35 & 0.0579920 & 0.0579916 \\ 0.40 & 0.0746422 & 0.0746408 \\ 0.45 & 0.0929963 & 0.0929918 \\ 0.50 & 0.1129240 & 0.1129120 \end{array} \right)$$ Now, for small values of $\alpha$, the Taylor expansion of $I$ gives $$I=\frac{1}{2} \alpha^2 - \frac{1}{4}\alpha^4 + \frac{1}{3}\alpha^6 - \frac{17}{24}\alpha^8 + \frac{31}{15}\alpha^{10} +O\left(\alpha^{12}\right)$$ as already given by Sangchul Lee.

Edit

We could contiue with $[2n+2,2]$ Padé approximants and get, after long division, $$\log(\cosh(y))=\sum_{k=0}^{2n} a_k y^{2k}+\frac{b}{1+c y^2}$$ leading to $$I=\sum_{k=0}^{2n} (2k-1)!!\,a_k\,\alpha^{2k+2} +\frac{\pi b}{\sqrt{c}}\,\frac{e^{\frac{1}{2 \alpha^2 c}}}{\alpha}\,\text{erfc}\left(\frac{1}{ \alpha \sqrt{2c}}\right)$$