Integration with a condition of constant values.

Question

I know that

$$\frac{1}{e^{\infty}}=0$$

Can I say that integration $$\int_0^{\infty} \frac{1}{e^{-ax}}dx=-a$$ where $a\in [0,-\infty)$. The reason is because the values of a is negative.

Is this true or I am wrong in this assumption.

Answer 1

\begin{align} \int_0^{\infty}e^{ax}\mathrm{d}x=\lim_{n\to \infty} \int_0^n e^{ax}\mathrm{d}x=\frac{1}{a}\lim_{n\to \infty}[e^{ax}+C]_0^n=\lim_{n\to \infty}\frac{e^{na}-1}{a} \end{align} which is $\infty$ is $a\ge 0$, and $-1/a$ if $a<0$.

Answer 2

$$\int_{0}^\infty \frac{1}{e^{-ax}}dx=\int_{0}^\infty e^{ax}dx$$

Which converges only for negative values of $a$. When it does converge, you have $$\int_{0}^\infty e^{ax}dx$$ and, since $a$ is negative, by using $u$-sub, we have $$=\frac{1}{a}\int_{0}^{-\infty} e^{u}du$$ $$=-\frac{1}{a}$$