If you are trying to evaluate \(\displaystyle \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx}\)

\(\displaystyle \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx} = \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t - \lambda)e^{x(i\,t - \lambda)}\,dx}\)

Now using integration by parts with

\(\displaystyle u = x\) so that \(\displaystyle du = 1\)

and \(\displaystyle dv = (i\,t - \lambda)e^{x(i\,t - \lambda)}\) so that \(\displaystyle v = e^{x(i\,t - \lambda)}\)

we find

\(\displaystyle \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t - \lambda)e^{x(i\,t - \lambda)}\,dx} = \left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \int{e^{x(i\,t - \lambda)}\,dx}\right]_0^{\varepsilon}\)

\(\displaystyle =\left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \frac{e^{x(i\,t - \lambda)}}{i\,t - \lambda}\right]_0^{\varepsilon}\)

Can you finish?