Taylor series of a definite integral Consider the function of a parameter $\alpha > 0$, given by
$$f(\alpha) = \frac{2}{\sqrt 2\pi} \int_0^\infty e^{\dfrac{-x^2}{2\alpha^2}}\cosh(x)\log\cosh(x) dx.$$
From Wolfram-alpha, it seems that for small values of $\alpha$, 
$$f(\alpha) = \frac{\alpha^3}{2}+\frac{\alpha^5}{2} + o(\alpha^5).$$
How can one establish this rigorously?
Edit: Per the comment below, I used dfrac to make it more visible.
 A: Considering $$f(\alpha) = \frac{2}{\sqrt {2\pi}} \int_0^\infty e^{\frac{-x^2}{2\alpha^2}}\cosh(x)\log\big(\cosh(x)\big) dx$$ start developing $\cosh(x)\log\big(\cosh(x)\big)$ as an infinite Taylor series built at $x=0$. We then get $$\cosh(x)\log\big(\cosh(x)\big)=\frac{x^2}{2}+\frac{x^4}{6}+\frac{x^6}{720}+\frac{x^8}{630}+O\left(x^9\right)$$ and so we are let with a linear combination of integrals of the type $$I_k=\int e^{\frac{-x^2}{2\alpha^2}}x^{2k}dx=-2^{k-\frac{1}{2}} \alpha ^{2 k+1} \Gamma \left(k+\frac{1}{2},\frac{x^2}{2 \alpha
   ^2}\right)=-\frac{1}{2} x^{2 k+1} E_{\frac{1}{2}-k}\left(\frac{x^2}{2 \alpha ^2}\right)$$ where appears the incomplete gamma function or the elliptic integral. 
Concerning the corresponding definite integral $$J_k=\int_0^{\infty} e^{\frac{-x^2}{2\alpha^2}}x^{2k}dx=2^{k-\frac{1}{2}} \left({\alpha ^2}\right)^{k+\frac{1}{2}} \Gamma
   \left(k+\frac{1}{2}\right)$$ So, replacing, we get $$f(\alpha)=\frac{\alpha ^3}{2}+\frac{\alpha ^5}{2}+\frac{\alpha ^7}{48}+\frac{\alpha
   ^9}{6}+\cdots$$
