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Chemistry LibreTexts

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:

    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

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