evaluate $\int_{0}^{1}\frac{x-1}{\ln x}dx$

Question

evaluate $\int_{0}^{1}\frac{x-1}{\ln x}dx$,where x is real.

Approach:

The suggestion is to differentiate $H(m)=\int_{0}^{1}\frac{x^{m}-1}{\ln x}dx$. This leads to

$$H(m)'=m\int_{0}^{1}\frac{x^{m-1}}{\ln x}dx, $$

$$H(m)''=\int_{0}^{1}\frac{x^{m-1}}{\ln x}dx+m(m-1)\int_{0}^{1} \frac{x^{m-2}}{\ln x}dx$$

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The notation suggests you need to be taking the derivative with respect to $m$ — David P, Feb 09 '14 at 23:20
See also: Prove $\int_{0}^1\frac{x-1}{\log x}dx=\log 2$ and $\int_{0}^1\frac{\log x}{x-1}dx=\frac{\pi^2}{6}$ and Why does $\int\limits_0^1 {\dfrac{{x - 1}}{{\ln x}}} ;\text{d}x=\ln2$? — Ben, Feb 09 '14 at 23:28

score 5 · Answer 1 · answered Feb 09 '14 at 23:21

5

The hint is very good. Let $$\eta(s)=\int_{0}^{1}\frac{x^s-1}{\log x}dx$$ Then $$\eta'(s)=\int_0^1 \log x\frac{x^{s}}{\log x}dx=\int_0^1 {x^{s}} dx$$

answered Feb 09 '14 at 23:21

Pedro

1 Answers1