Integration (by parts?)

Question

I don't know how the answer is obtained in the following.

$2\lambda^2 \int_{0}^{\infty}ye^{-\lambda(1+z)y}dy = \frac{2}{(1+z)^2}\int_{0}^{\infty}ue^{-u}dy = \frac{2}{(1+z)^2}, 0<z < 1$

I tried to solve it by integration by parts, and can't seem to get it.

$$u = y \Rightarrow du = dy$$ $$dv = e^{-\lambda(1+z)y} \Rightarrow v = \int_{0}^{\infty}e^{-\lambda(1+z)y} dy = \frac{1}{\lambda(1+z)}$$ $$2\lambda^2 \int_{0}^{\infty}ye^{-\lambda(1+z)y}dy = 2\lambda^2\{\frac{y}{\lambda(1+z)} \bigg|_{0}^{\infty} - \int_{0}^{\infty}\frac{1}{\lambda(1+z)}du\} = 2\lambda^2\{\frac{y}{\lambda(1+z)} \bigg|_{0}^{\infty} - \frac{u}{\lambda(1+z)} \bigg|_{0}^{\infty}\}$$

Answer 1

They made a substitution. We have:

$$2\lambda^{2}\int ye^{-\lambda(1+z)y}\text{d}y$$

Let's substitute $u=\lambda(1+z)y$, so that " $\text{d}u=\lambda(1+z)\text{d}y$ ". We have $y=\frac{u}{\lambda(1+z)}$ and $\text{d}y=\frac{\text{d}u}{\lambda(1+z)}$

$$\begin{align}2\lambda^{2}\int \frac{u}{\lambda(1+z)}e^{-u}\frac{\text{d}u}{\lambda(1+z)} &= \frac{2\lambda^{2}}{\lambda^{2}(1+z)^{2}}\int ue^{-u}\text{d}u\\ &=\frac{2}{(1+z)^{2}} \left(-ue^{-u}-\int -e^{-u}\text{d}u\right)\\ &=\frac{2}{(1+z)^{2}}(-ue^{-u}-e^{-u})\\ &=-\frac{2}{(1+z)^{2}}e^{-u}(u+1) \end{align}$$

and

$$2\lambda^{2}\int ye^{-\lambda(1+z)y}\text{d}y=\frac{-2}{(1+z)^{2}}\left[e^{-\lambda(1+z)y}(\lambda(1+z)y+1)\right]$$

and the corresponding definite integral is as follows:

$$\begin{align}2\lambda^{2}\int_{0}^{\infty} ye^{-\lambda(1+z)y}\text{d}y &=\frac{-2}{(1+z)^{2}}\left[e^{-\lambda(1+z)y}(\lambda(1+z)y+1)\right]_{0}^{\infty}\\ &=0-\frac{-2}{(1+z)^{2}}\left[e^{-0}(\lambda(1+z)0+1)\right]\\ &=\frac{2}{(1+z)^{2}} \end{align}$$

Be aware that I assumed $\lambda>0$ and $0<z<1$. If $\lambda<0$, the integral is divergent.

Answer 2

The $v$ is not right, it should be $v=-\frac{1}{\lambda(1+z)}e^{-\lambda (1+z)y}$. We want an antiderivative of $e^{-\lambda (1+z)y}$.