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How do I evaluate $I= \int_0^1 \frac{x+1}{\log x} dx$?

This seemed easy at the first glance, but I could not solve this. Is there an elementary way to evaluate this integral?

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    $\begingroup$ There is no elementary antiderivative to that integrand. $\endgroup$ – Alex Nov 19 '15 at 5:13
  • $\begingroup$ @Alex I think so too since it is a sum of Ei(x)'s. However, my friend asked me this problem and he said it was on the graduate entrance exam. So I tried to apply convergence theorems which are taught in undergraduate course, but it does not work. $\endgroup$ – Rubertos Nov 19 '15 at 5:16
  • $\begingroup$ @ Alex: True, but after substitution, you might end up with an integral like $\int_0^\infty e^{-u^2}\,du$, which can be evaluated exactly. (The fact that there are limits suggests that you might be able to evaluate the integral without finding an explicit antiderivative.) $\endgroup$ – Christopher Carl Heckman Nov 19 '15 at 5:29
  • $\begingroup$ @CarlHeckman What substitution would make that work $\endgroup$ – Rubertos Nov 19 '15 at 5:43
  • $\begingroup$ Maple and WolframAlpha say that the value is $-\infty$. You might be able to get it by letting $x=e^{-y}$ and then doing a power series analysis. $\endgroup$ – Christopher Carl Heckman Nov 19 '15 at 5:45
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I assume you meant $~I=\displaystyle\int_0^1\frac{x\color{red}-1}{\log x}~dx,~$ since $~\displaystyle\int_0^1\frac{x\color{red}+1}{\log x}~dx~$ clearly diverges in $1$. The former can be evaluated by differentiation under the integral sign, as Juan Theron's now deleted answer so beautifully shows it. Alternately, one could also use $x=e^t$, and then employ Frullani's integral, which is based on expressing the integrand itself as a second integral, and reversing the order of integration.

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