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I understand that integrating (x-1)/ln(x) is a tricky task in the general case, but I am hoping that restricting the problem to a definite integral over a non-negative domain simplifies the answer, to the point that it can be expressed in a closed form that doesn't include gamma functions, dilogarithms, Ei, or any other concept that someone shouldn't be expected to remember 30 barren years after last taking calculus!

Thus, is there such a solution to: $\int_{0}^w \frac{x-1}{ln\ x}\ dx$

If not, is there such a solution to: $\int_{1}^w \frac{x-1}{ln\ x}\ dx$

Thank you.

Iron Pillow
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2 Answers2

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The simple answer is no, there is no closed form of this integral, because it results in the exponential and logarithmic integral, which have no closed forms. See the general antiderivative:

$$ \int \frac{x-1}{\ln x}dx = \text{Ei}(2\ln x) - \text{li}(x) +c $$

Edit: In case you're interested, Mathematica comes up with the following for the latter of your integrals,

$$ \int_1^w \frac{x-1}{\ln x}dx = -\ln(-2\ln w) + \ln(-\ln w) - \Gamma(0,-2\ln w) + \Gamma(0,-\ln w), \quad \Im [w]\ne 0 \lor\Re [w]>0 $$

where $\Gamma(a, b)$ is the incomplete gamma function.

Dando18
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  • Can you explain the notations at the end of the definite integral expression? For my situation, w>e, so that term ln(-ln w) is going to blow up. I'm guessing that the notations restrict to w<e, but I just want to make sure. – Iron Pillow Jul 27 '17 at 17:13
  • @IronPillow $\Im[w] \ne 0 \ \lor\ \Re[w]>0 $ means the imaginary part of $w$ is not $0$ or the real part of $w$ is greater than $0$. Now this is assuming $w\in\mathbb C$, however I get the feeling you have restricted $w\in\mathbb R$. Which means $\Im[w]=0$ so $\Re[w]$ must be greater than zero. Basically $w$ must be greater than $0$ for the equality to hold, if you're considering $w$ as a real number. – Dando18 Jul 27 '17 at 17:22
  • @IronPillow are you trying to find a good numerical approximation? or are you simply interested in a closed form solution? – Dando18 Jul 27 '17 at 17:27
  • Thank you for the clarification. You are correct: w is real. I am looking for a closed-form solution. I suspect the gamma functions will allow a numerical computation, but I still don't understand the term ln(-ln w)... if w=$e^2$, for example, then we have ln(-ln $e^2$) = ln(-2). So I must be missing something here, too. – Iron Pillow Jul 27 '17 at 17:45
  • @IronPillow Use the complex definition of the logarithm and the fact that $\ln a + \ln b = \ln(ab)$ is true for complex $a$ and $b$. So you'll end up getting $\ln(-2) + \ln(-4) = \ln((-2)(-4)) = \ln8$. So you end up with a real result. And the complex parts of the incomplete gamma functions cancel out in the same manner. – Dando18 Jul 27 '17 at 18:06
  • Brilliant! Thank you. – Iron Pillow Jul 27 '17 at 21:29
  • @IronPillow Consider marking it correct if it helped you :) – Dando18 Jul 28 '17 at 00:45
  • Will do. I am getting greedy now and am hoping for a pointer to a derivation of your answer. :-) – Iron Pillow Jul 28 '17 at 14:01
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$\int\frac{x-1}{\ln x}\,\mathrm{d}x$ cannot be expressed in terms of elementary functions, however $I(\alpha)=\int_{0}^{1}\frac{x^\alpha-1}{\ln x}\,\mathrm{d}x$ can be by partial differentiation under the integral. A good explanation of this can be found here. You get that $I(\alpha)=\ln(\alpha+1)$

Teh Rod
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  • It's tantalizing. Integration over 0-to-1 produces such a clean answer, but integration over 0-to-w (w>1) is ugly. :-( – Iron Pillow Jul 27 '17 at 17:08