# Exponential Integrals

- Page ID
- 5554

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: |
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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

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