How can I integrate $\int_{x}^{\infty}\lambda ye^{-\lambda(y-x)}\mathrm{d}y$ by parts

Question

Conditional expectation can be calculated using integration:

$$\int_{x}^{\infty}yf_{Y|X}(y|x)\mathrm{d}y = \int_{x}^{\infty}\lambda ye^{-\lambda(y-x)}\mathrm{d}y = x + \frac{1}{\lambda}\quad\text{for}\quad x\geq 0$$

My question is how to show $x + \frac{1}{\lambda}\quad\text{for}\quad x\geq 0$

Answer 1

Notice that you are integrating over $y$. So you can factor the terms which do no not depend on it.

Precisely, we have that \begin{align*} \int_{x}^{\infty}\lambda y\exp(-\lambda(y-x))\mathrm{d}y & = \exp(\lambda x)\int_{x}^{\infty}\lambda y\exp(-\lambda y)\mathrm{d}y\\\\ \end{align*}

The last one can be computed through integration by parts. Indeed, it results that \begin{align*} \int_{x}^{\infty}\lambda y\exp(-\lambda y)\mathrm{d}y & = -y\exp(-\lambda y)\Biggr|_{x}^{\infty} + \int_{x}^{\infty}\exp(-\lambda y)\mathrm{d}y\\\\ & = -y\exp(-\lambda y)\Biggr|_{x}^{\infty} - \frac{\exp(-\lambda y)}{\lambda}\Biggr|_{x}^{\infty}\\\\ & = x\exp(-\lambda x) + \frac{\exp(-\lambda x)}{\lambda} \end{align*}

Consequently, the proposed integral reduces to \begin{align*} \int_{x}^{\infty}\lambda y\exp(-\lambda(y-x))\mathrm{d}y & = \exp(\lambda x)\left(x\exp(-\lambda x) + \frac{\exp(-\lambda x)}{\lambda}\right) = x + \frac{1}{\lambda} \end{align*}

just as desired.