I have the following integral:
$$ \frac{1}{\gamma^2}\int_0^\infty \exp\left(-\frac{t}{\gamma}-\frac{|t-\tau|}{\mu}\right) \left(t-\frac{t^2}{2\gamma}\right)dt = \frac{\gamma/\mu}{(1-\gamma/\mu)^3}e^{-\tau/\mu} $$
for any $\tau>0$. The solution reflects that the integral blows up when $\gamma = \mu$.
Could someone explain why? What is the mechanism of the blow-up?
You must consider to solve: $$I=\int_{0}^{\infty}\exp{\left(-\frac{t}{\gamma}-\frac{|t-\tau|}{\mu}\right)}\left(t-\frac{t^2}{2\gamma}\right)dt=\int_{0}^{\tau}\exp{\left(-\frac{t}{\gamma}-\frac{\tau-t}{\mu}\right)}\left(t-\frac{t^2}{2\gamma}\right)dt+\int_{\tau}^{\infty}\exp{\left(-\frac{t}{\gamma}-\frac{t-\tau}{\mu}\right)}\left(t-\frac{t^2}{2\gamma}\right)dt$$ You see that the first and second integrals are convergent even if $\gamma=\mu>0$
For the case $\mu=\gamma=\frac{1}{\eta}$ $$I=\exp{(-\eta\tau)}\left(\tau^2+\frac{\tau^2}{3}\eta\right)-\frac{1}{4\eta^2}\frac{d}{d\alpha}\left[\frac{\exp{(-2\tau\eta\alpha)}}{\alpha}+\frac{1}{4}\frac{d}{d\alpha}\left(\frac{\exp{(-2\tau\eta\alpha)}}{\alpha}\right)\right]_{\alpha=1}$$ Hope I did not any mistake.