how can I solve this integral$ \int_{0}^{\infty}\frac{(v - \lambda)^2}\lambda\exp(\frac {\text -v}{\lambda}) \,\mathrm dv$ $$\int_{0}^{\infty}\frac{(v - \lambda)^2}\lambda\exp(\frac {\text -v}{\lambda})
\,\mathrm dv$$
$$t=\frac {\text -v}{\lambda}$$
$$dt=\frac {\text -dv}{\lambda}$$
$$=-\lambda\int_{-\infty}^{0}\frac{(-\lambda t - \lambda)^2}\lambda\exp(t)
\,\mathrm dt$$
How should I do to continue?
 A: First, it seems like, if you are trying to evaluate $\int\limits_{-\infty}^{0}\frac{(v-\lambda)^2}{\lambda}\cdot e^{\frac{-v}{\lambda}}\partial v$, then the answer is that it simply diverges, as well, as if the bounds for integration were $(-\infty, +\infty)$, as $e^{\frac{-v}{\lambda}} \to +\infty,~\text{as }v \to -\infty$. Moreover, in your $t$-substitution, you have forgotten to change the integration bounds.
Now, let's assume, that you are looking for $\int\limits_{0}^{+\infty}\frac{(v-\lambda)^2}{\lambda}\cdot e^{\frac{-v}{\lambda}}\partial v$. Then, if (as correctly mentioned by @RAHUL)$\lambda \in \mathbb{R}^+$:
$$
\int\limits_{0}^{+\infty}\frac{(v-\lambda)^2}{\lambda}\cdot e^{\frac{-v}{\lambda}}\partial v~=~\left[t = \frac{-v}{\lambda},~\partial t = -\frac{\partial v}{\lambda} \biggm\vert~v = -\lambda t,~\partial v = -\lambda\partial t\right] = \\=(\text{Since we swap zero to be the upper bound, we have to change the sign}) = -1\cdot(-\lambda)\cdot\\\cdot\int\limits_{-\infty}^{0}\frac{(-\lambda t-\lambda)^2}{\lambda}\cdot e^{t}\partial t = \lambda^2 \int\limits_{-\infty}^{0}(t+1)^2 e^{t}\partial t = \lambda^2 \cdot\\\cdot\left(\int\limits_{-\infty}^{0}e^t \cdot t^2~\partial t + 2\int\limits_{-\infty}^{0}e^t \cdot t~\partial t + \int\limits_{-\infty}^{0}e^t\partial t\right) = \left[\text{Integrating by parts}\right] = \\=\lambda^2\left(e^tt^2\biggm\vert^{0}_{\lim\limits_{n\to-\infty}(n)} - \int\limits_{-\infty}^{0}e^t \cdot 2t~\partial t + 2\int\limits_{-\infty}^{0}e^t \cdot t~\partial t \int\limits_{-\infty}^{0}e^t\partial t \right) =\\= [\text{As }e^t \text{ has higher convergence rate to }0\text{ then the divergence rate of } t^2 \text{ at }-\infty]=\\=\lambda^2\left([0 - 0] + \int\limits_{-\infty}^0 e^t \partial t \right) = \lambda^2 \cdot \lim\limits_{n\to-\infty}\left(e^t\biggm\vert^{0}_{n}\right) = \lambda^2 \cdot (1 - 0) = \lambda^2.
$$
