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Explain why ex / x cannot be integrated by parts

Try to use integration by parts to evaluate the integral

    \[ \int \frac{e^x}{x} \, dx. \]

If we let

    \begin{align*}  u &= \frac{1}{x} & \implies && du &= -\frac{1}{x^2} \, dx \\ dv &= e^x \, dx& \implies && v &= e^x, \end{align*}

Then we have

    \[ \int \frac{e^x}{x} \, dx = \frac{e^x}{x} + \int \frac{e^x}{x^2} \, dx. \]

If we try to integrate by parts again with u = \frac{1}{x^2} and dv = e^x then we’ll end up with another integral, this time with \frac{e^x}{x^3}. That will continue, so we’ll just keep getting integrals that we can’t evaluate.

On the other hand, if we try letting

    \begin{align*}  u &= e^x & \implies && du &= e^x \, dx \\ dv &= \frac{1}{x} \, & \implies && v &= \log x, \end{align*}

Then we have

    \[ \int \frac{e^x}{x} \, dx = e^x \log x - \int e^x \log x \, dx. \]

Continuing along that route, we’ll keep getting integrals of x^k e^x \log x where k keeps getting larger.

In both cases, we keep getting increasing complicated integrals that we can never evaluate.


  1. Anonymous says:

    Perhaps the solution could be expersee
    \int\frac{e^{x}}{x} = \frac{e^{x}}{x} + \int \frac{e^{x}}{x^2} = \frac{e^{x}}{x} + \frac{e^{x}}{x^2} + 2 \int \frac{e^{x}}{x^3} = \dots = e^x\sum\limits_{k=1}^{\infty} (k-1)! x^{-k}

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