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$\int_0^1 \frac{x-1}{\ln(x)} dx = \ln(2)$

First i try $\ln(x)=t$ so that $\frac{1}{x} dx =dt$ then integral becomes \begin{align} &\int_{-\infty}^{0}\frac{e^t-1}{t} (e^t dt) = - \int_0^{\infty} \frac{e^{-t}-1}{t} e^{-t} dt = -\int_0^{\infty} (t^{-1} e^{-2t} - t^{-1} e^{-t})dt \\ & = - \ln(t) e^{-2t} |_{0, \infty} - 2 \int_{0}^{\infty} \ln(t) e^{-2t}dt + \ln(t) e^{-t}|_{0, \infty} + \int_0^{\infty} \ln(t) e^{-t} dt \end{align} This substitution reproduce the integral form of \begin{align} \int_0^\infty \ln(t) e^{-t} dt \end{align}

hmm.

I want to know other kinds of evaluating $\int_0^1 \frac{x-1}{\ln(x)} dx = \ln(2)$ and evaluating of $\int_0^\infty \ln(t) e^{-t} dt$.

phy_math
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    Doesn't $\displaystyle{-\int_{0}^{\infty} \ln(t) e^{-2t}d(2t)= -\int_{0}^{\infty} (\ln(2t)-\ln 2) e^{-2t}d(2t)}$ and $\displaystyle{\int_{0}^{\infty} \ln(t) e^{-t}dt}$ cancel to make $\ln 2$ ? – Alexey Burdin May 20 '15 at 06:23
  • @AlexeyBurdin, Oh Thanks! – phy_math May 20 '15 at 15:39

2 Answers2

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Let $I(b) = \int\limits_0^1 \dfrac{x^b - 1}{\ln x} dx$. Then : \begin{align} I'(b) &= \int\limits_0^1 \dfrac{\partial }{\partial b}\dfrac{x^b - 1}{\ln x} dx\\ &=\int\limits_0^1 x^b dx\\ &= \dfrac{1}{b+1} \end{align} Hence : $$I(b) = \ln (b+1) + C$$ with $C\in \mathbb{R}.$ To find $C$ note that $I(0) = 0$ which implies $C= 0$. So $I(1) = \ln(2)$. To handle the second integral note that : $$\Gamma (x)= \int\limits_0^\infty t^{x-1}e^{-t}dt,\qquad \Gamma' (x)= \int\limits_0^\infty t^{x-1}e^{-t}\ln tdt$$ As shown here, since $\Gamma$ is log-convex one has : $$\dfrac{\Gamma'(x)}{\Gamma(x)} = -\gamma + \sum_{k=0}^\infty \dfrac{1}{k} - \dfrac{1}{k+x-1}.$$ So $$\int\limits_0^\infty \log(t) e^{-t} dt = \Gamma'(1) = -\gamma$$ where $\gamma$ is the Euler–Mascheroni constant.

Community
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Joelafrite
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  • Thanks for detailed procedure! I think i am well aware of gamma function and digamma function, but can not think of using here in this integral!. Your answer enlarge my world! – phy_math May 20 '15 at 15:41
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Sometimes, the best way of simplifying something is by making it more complicated. :-$)$

Hint: Evaluate $I(k)=\displaystyle\int_0^1\frac{x^k-1}{\ln x}~dx$ by differentiating under the integral sign with regards to k.

Lucian
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