Consider the following improper integral$$f(\mu, t) = \frac{2}{\pi} \int_0^\infty \frac{\sin u}{u} \frac{\left(U+u\right) \exp\left(-\frac{\mu}{t} \, u\right) - 2U \exp\left(-\frac{\mu}{t} \, U \right)}{U-u} \, \mathrm{d} u \, , $$where $U = (u^2 + t^2)^{\frac{1}{2}}$ with $\mu, t \ge 0$.Here, I am interested in the asymptotic behavior of this integral as $t \to \infty$.
By taking the limit $t\to\infty$, it can easily be shown that
$$\lim_{t \to \infty} f(\mu, t) = 1 - 2\exp(-\mu) \, .$$However, I would like to understand how the function approaches this limit. Any insights or assistance would be greatly appreciated.
It is important to note that expanding the function in a series around $t \to \infty$ and then integrating with respect to $u$ results in an undefined integral.Specifically, the integrand can be expanded around $t=\infty$ as$$\frac{2}{\pi} \, \sin u\left( \frac{1-2e^{-\mu}}{u} + \frac{\mu}{t} \left( 2-\mu-2e^{-\mu} \right) \right) + \mathcal{O}(t^{-2}) \, .$$
Thanks!