An integral with $\frac{1}{\log x}$ [Solution]

The problem was:

Suppose $n$ is a positive integer (though the result below does not really need that).

Show that

$$\int_{0}^{1} \frac{x^n - 1}{\log x} \text{d}x = \log(n+1)$$

Note that the $\log x$ is the $\log$ to base $e$.

Solution

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This can be solved by the neat trick of differentiating under the integral sign.


Let

  $$f(z) = \int_{0}^{1} \frac{x^z - 1}{\log x} \text{d}x$$

Differentiating under the integral sign gives us

  $$f'(z) = \int_{0}^{1} \frac{d \frac{x^z - 1}{\log x}}{dz} \text{d}x$$

and so

  $$f'(z) = \int_{0}^{1} \frac{x^z \log x}{\log x} \text{d}x  = \int_{0}^{1} x^z \text{d}x  = \frac{1}{z+1}$$

Thus $f(z) = \log(1 + z)$, since $f(0) = 0$.

[Note: There was some handwaving and the right theorems need to be applied, and right bounds on $z$ need to be assumed etc. That is left to the reader]