I wish to calculate $$\int_{0}^{\infty}dx\int_{0}^{xz}\lambda^{2}e^{-\lambda(x+y)}dy $$
I compared my result, and the result with Wolfram when setting $\lambda=3$ and I get different results.
What I did:
$$\int_{0}^{\infty}dx\int_{0}^{xz}\lambda^{2}e^{-\lambda(x+y)}dy $$
$$=\lambda^{2}\int_{0}^{\infty}\frac{e^{-\lambda(x+y)}}{-\lambda}|_{0}^{xz}\, dx$$
$$=\lambda^{2}\int_{0}^{\infty}\frac{e^{-\lambda(x+xz)}}{-\lambda}-\frac{e^{-\lambda x}}{-\lambda}\, dx$$
$$=-\lambda(\int_{0}^{\infty}e^{-\lambda(1+z)x}\, dx-\int_{0}^{\infty}e^{-\lambda x}\, dx)$$
$$=-\lambda(\frac{e^{-\lambda(1+z)x}}{-\lambda(1+z)}|_{0}^{\infty}-\frac{e^{-\lambda x}}{-\lambda}|_{0}^{\infty})$$
$$=-\lambda(\frac{1}{\lambda(1+z)}-(\frac{1}{-\lambda}))$$
$$=-\lambda(\frac{1}{\lambda(1+z)}+\frac{1}{\lambda})$$
$$=1-\frac{1}{1+z}$$
I went over the calculation a couple of times and not only that I couldn't find my mistake, I also don't understand how I can end up with $1-e^{\text{something}}$ because the integrals are done from $0$ to $\infty$ and then I get $1$ or $0$ when I set the limits.
Can someone please help me understand where I am mistaken ?