Asymptotic expansion of an integral with exponent times another function

Question

According to this book page number 192

The asymptotic expansion of

$$\int_{0}^{\infty}[1-t \log t+(1-C-2 \log 2) t] e^{-2 t \log \left(\mu H / T_{1}\right)} d t$$

is

$$\frac{1}{2 \log \frac{\mu H}{T_{1}}}-\frac{\log 2}{\left(2 \log \frac{\mu H}{T_{1}}\right)^{2}}+\frac{\log \log \frac{\mu H}{T_{1}}}{\left(2 \log \frac{\mu H}{T_{1}}\right)^{2}}+\cdots$$

How is this computed?

Here is my try

The integrand can be written as

$$e^{{-2 t \log \left(\mu H / T_{1}\right)}+\log\left([1-t \log t+(1-C-2 \log 2) t] \right)}$$

Now, most of the contribution comes from the extremum of the function which is given by

$$0=\frac{-C-\log (t)-2 \log (2)}{t (-C+1-2 \log (2))-t \log (t)+1}-2 \log \left(\frac{H \mu }{\text{T1}}\right)$$

Now, I do not know how to invert this relation to find an expression for $t_{ext}$.

Is there any other method to calculate the asymptotes other than the steepest decent method?

Answer 1

Assuming that $\Re(\lambda )>0$, we have $$\int_0^\infty (a t-t \log (t)+1)\,e^{-\lambda t}=\frac{\lambda +\log (\lambda )+(a+\gamma -1)}{\lambda ^2}$$ which is exact and gives your expression.

Answer 2

I think splitting it would do the work. The formula is just the question of calculating something like:
$ \int_{0}^{\infty} (1- t\log t +at) e^{-\lambda t}dt$
After splitting, each term is .
i) $\int_{0}^{\infty} e^{-\lambda t} dt= \frac{1}{\lambda}$
ii)$\int_{0}^{\infty} te^{-\lambda t} dt= \frac{1}{\lambda^2}$
iii)$\int_{0}^{\infty} t\log t .e^{-\lambda t}dt =\frac{1}{\lambda} \int_{0}^{\infty}( 1+\log t)e^{-\lambda t}dt $
For the last term, I don't know how to control but I guess as a physcist, you're provided with enough approximation formula to solve this term better than me.