Complex integration?

Calculus Level 5

$\large \int_0^1 \ln(1-e^x) \, dx$

If we consider the branch of the complex logarithm to be $\ln(-x) = \ln x + i \pi$ , then the complex integral above is equal to $\text{Li}_a (e^{-b}) + \dfrac cd + f \cdot i\pi - \zeta (g),$ where $a,b,c,d,f,g$ are positive integers with $c,d$ coprime, find $a+b+c+d+f+g$ .

Notations :

${ \text{Li} }_{ n }(a)$ denotes the polylogarithm function, ${ \text{Li} }_{ n }(a)=\displaystyle\sum _{ k=1 }^{ \infty }{ \frac { { a }^{ k } }{ { k }^{ n } } }.$ "
$i = \sqrt{-1}$ .
$\zeta(\cdot)$ denotes the Riemann zeta function .

The answer is 9.

1 solution

Mark Hennings
Apr 21, 2016

Assuming that we are working with a suitable branch of the logarithm such that $\log(-x) = \ln x + i\pi$ for $x > 0$ , we have $\begin{array}{rcl} \displaystyle \int_0^1 \log(1-e^x)\,dx & = & \displaystyle \int_0^1 \big[\ln (e^x - 1) + i\pi\big]\,dx \; = \; \int_0^1 \big[x + \ln(1 - e^{-x})\big]\,dx + i\pi \\ & = & \displaystyle \tfrac12 - \int_0^1 \sum_{n=1}^\infty \frac{1}{n} e^{-nx}\,dx + i\pi \; =\; \tfrac12 - \sum_{n=1}^\infty \frac{1}{n^2}\big[1 - e^{-n}\big] + i\pi \\ & = & \displaystyle \tfrac12 - \zeta(2) + \mathrm{Li}_2(e^{-1}) + i\pi \end{array}$ making the answer $2+1+1+2+1+2 = \boxed{9}$ .

Complex integration?

The answer is 9.

1 solution

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