# Exponential Integrals

Exponential Function $e^{-ax}$

The area under the curve is:

$\int_0^{\infty} e^{-ax} \,dx = \dfrac{1}{a}$

To obtain this answer use the substitution $$u = -ax$$ and $$du = -a\,dx$$. Therefore $$dx = -du / a$$.

The integral can be written:

$-\frac{1}{a}\int_{0}^{\infty}\,e^u\,du=-\frac{1}{a}\,e^u\Biggr\rvert_{0}^{\infty}=-\frac{1}{a}(0-1)=\frac{1}{a}$

In general exponential integrals multiplied by a polynomial can be solved using integration by parts.

Example 1:

For example, the integral

$\int_0^\infty x e^{-ax}\;dx$

can be solved by saying that $$u = x$$ and $$dv = e^{-ax} \,dx$$. We can use

du = dx and v = (-1/a)e-ax to obtain the integral. In general

$\int u\,dv=\ uv-\int v\,du$

so that in the present case we have

$\int_{0}^{\infty}\,xe^{-ax}\,dx=\frac{1}{a}\,xe^{-ax}\Biggr\rvert_{0}^{\infty}+\frac{1}{a}\int_{0}^{\infty}\,e^{-ax}\,dx=\frac{1}{a^2}$

Notice that the first ($$uv$$) term is zero and the second term contains the integral of $$e^{-ax}$$ that was solved above.

By this technique any polynomial can be solved yielding the general formula

$\int_{0}^{\infty}\,x^{n}\,e^{-ax}\,dx=\frac{n!}{a^{n+1}}$