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I want to calculate following definite integral:

$$\int\limits_{-1}^0\left(\frac{1}{x\ln(1 + x)}-\frac{1}{x^2}-\frac{1}{2x}\right)dx$$

No idea from where to begin.

I know result is:

$$-\frac{1}{2}-\frac{\gamma_0}{2}+\frac{\ln(2\pi)}{2}$$

an4s
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24th_moonshine
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  • I'd start by replacing $x$ by $-x$, so I'm integrating over positive values of $x$. – Angina Seng Jan 05 '20 at 08:57
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    $\int 1/x\log(1+x) dx$ has no elementary primitive in terms of standard functions https://www.wolframalpha.com/input/?i=integral+1%2F%28xlog%281%2Bx%29%29 – Gabrielek Jan 05 '20 at 09:02
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    Do oyu require an exact analytic expression or will an approximation suffice? If the latter, what are your precision and accuracy requirements? – Eric Towers Jan 05 '20 at 10:19
  • Best shot is to expanded the reciprocal log in a Laurent series... – Allawonder Jan 05 '20 at 13:32

1 Answers1

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The formula to be proved is (this corrects a sign error in the original statment) $$ \int_{-1}^0 \frac{dx}{x^2}\Big( \, \frac{x}{\log{(1+x)}} -1 -\frac{x}{2} \Big) = \frac{1}{2}\big(1+\gamma - \log{(2 \pi)} \big).$$

The expression within the parentheses within the integral can be written in terms of the Bernoulli polynomials of the second kind (see the wiki)

$$ \frac{x}{\log{(1+x)}} -1 -\frac{x}{2} = \sum_{n=2}^\infty a_n \, x^n \quad,\quad a_n = \int_{0}^1 \binom{u}{n} du.$$

Yes, that is a binomial within the integral. Also, $a_0=1, a_1=1/2.$ The radius of convergence is $|x|<1$ and since the integrand no longer has any problems within its domain, we can switch integral with sum and integrate right on up to -1.

$$ \int_{-1}^0 \frac{dx}{x^2}\Big( \, \frac{x}{\log{(1+x)}} -1 -\frac{x}{2} \Big) = \sum_{n=0}^\infty a_{n+2} \int_{-1}^0 x^n\, \frac{dx}{x^2} $$

$$ =\sum_{n=0}^\infty a_{n+2}\frac{(-1)^n}{n+1} = \int_0^1 du \sum_{n=0}^\infty \binom{u}{n+2}\frac{(-1)^n}{n+1} $$ $$ = \int_0^1 du \, u\big(-1+\gamma + \psi(1+u) \big) $$ $$ = \frac{1}{2}\big(1+\gamma - \log{(2\pi)} \big)$$ In the penultimate line and last line I let Mathematica do the work for me, although I think the series and integral, which involves the polygamma function $\psi$, have been collected in tables of integral and series before.

user321120
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