Calculate $\sum\limits_{n=0}^{\infty}{\int\limits_{\frac{1}{2}}^{\infty}(1-e^{-t})^{n}e^{-t^2}dt}$

$$\mbox{Calculate}\quad \sum_{n = 0}^{\infty}\int_{1/2}^{\infty} \left(1 - {\rm e}^{-t}\right)^{n}{\rm e}^{-t^{2}}{\rm d}t$$

• Basically I don't know where to start.
• I was thinking of using Tonelli Theorem but I have no idea how to calculate this sum, and neither do I know how to solve this integral.
• I also tried to expand $$\left(1 - {\rm e}^{-t}\right)^{n}$$ but that seems to lead nowhere.

This is a problem from Introductory Measure Theory course so I was expecting some of those methods to work. Any help would be greatly appreciated.

• You do not need to have an infinite upper bound since, after completinf the square $$\int_a ^b e^{t-t^2}\,dt=\frac{1}{2} \sqrt[4]{e} \sqrt{\pi } \left(\text{erf}\left(\frac{1}{2}-a\right)-\text{erf}\left(\frac{1}{2}-b\right)\right)$$ Commented May 28 at 6:17

\begin{align*} \sum_{n = 0}^\infty \int_{\frac{1}{2}}^\infty (1 - e^{-t})^n e^{-t^2}dt &= \int_{\frac{1}{2}}^\infty \sum_{n = 0}^\infty (1 - e^{-t})^n e^{-t^2}dt \end{align*}
Recall the formula $$\sum_{n = 0}^\infty q^n = \dfrac{1}{1 - q} \ \forall q, \vert q \vert < 1$$ Since $$0 < (1 - e^{-t}) < 1 \ \forall t \ge \dfrac{1}{2}$$, we have $$\sum_{n = 0}^\infty (1 - e^{-t})^n e^{-t^2} = \dfrac{e^{-t^2}}{1 - (1 - e^{-t})} = e^{t - t^2}$$ Thus, $$\sum_{n = 0}^\infty \int_{\frac{1}{2}}^\infty (1 - e^{-t})^n e^{-t^2}dt = \int_{\frac{1}{2}}^\infty \exp(t - t^2)dt$$
Notice that $$\exp(t - t^2) = \exp\left(-\left(t - \dfrac{1}{2}\right)^2\right)\exp(1/4)$$ and $$\int_{1/2}^\infty \exp\left(-\left(t - \dfrac{1}{2}\right)^2\right)dt = \int_0^\infty \exp(-u^2)du = \dfrac{\sqrt{\pi}}{2}$$
Therefore, $$\sum_{n = 0}^\infty \int_{\frac{1}{2}}^\infty (1 - e^{-t})^n e^{-t^2}dt = \dfrac{\sqrt{\pi}e^{1/4}}{2}$$
• And, if I am correct, we can interchange summation and integration because $(1-e^{-t})^{n}e^{-t^2}$ is always positive? Commented May 28 at 2:28