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### Evaluation of $\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx$ [duplicate]

Evaluation of $$\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx$$ Let $$I = \int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx = \int^{1}_{0}\frac{(1-x)+\ln x}{(1-x)\ln x}dx$$ Now How ...
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### Computing $\int_0^\infty \frac{\log x}{\exp x} \ dx$ [duplicate]

I know that $$\int_0^\infty \frac{\log x}{\exp x} = -\gamma$$ where $\gamma$ is the Euler-Mascheroni constant, but I have no idea how to prove this. The series definition of $\gamma$ leads me ...
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### Integral representation of Euler's constant

Prove that : $$\gamma=-\int_0^{1}\ln \ln \left ( \frac{1}{x} \right) \ \mathrm{d}x.$$ where $\gamma$ is Euler's constant ($\gamma \approx 0.57721$). This integral was mentioned in Wikipedia as in ...
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### Evaluating $\int_0^\infty (\operatorname{E}_n(x)e^x-\frac1{1+x})dx$

I want to evaluate $$I_n=\int_0^\infty \left(\operatorname{E}_n(x)e^x-\frac1{1+x}\right)dx=-\psi(n)$$where $\operatorname{E}_n$ denotes exponential integral and $\psi$ denotes polygamma function. ...
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### Integral representation of Euler-Mascheroni constant

I'm solving an exercise in complex analysis. I proved $\displaystyle \gamma=\int_0^1 -\log(\log(1/s)) \, ds$. I want to prove that $\displaystyle \gamma=\int_0^1 \frac{1-e^{-t}-e^{-1/t}} t \,dt$. ...
### How to solve the integral $I=\int_0^\infty (e^{-u}\log u) du$ using standard methods [duplicate]
The following question from $\textit{Statistical Inference }$by Casella & Berger was given to me on an assignment last semester. On the surface, it appeared to be a straight-forward question about ...