Change in normal integrals

Calculus Level 5

$\large \displaystyle\int _{ 0 }^{ 1 }{ \ln { \left( 1-{ e }^{ \sqrt { x } } \right) \, dx } }$ The above integral is in the form: $a\left[\text{Li}_b(e^{-c})+\text{Li}_d(e^{-f}) -\zeta (g)+\frac{1}{h} \right]+i\pi,$

where $a,b,c,d,f,g$ and $h$ are positive integers. Find $a+b+c+d+f+g+h-2$ .

Notations :

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

The answer is 13.

1 solution

Mark Hennings
Apr 22, 2016

Working with an appropriate branch of the complex logarithm, $\begin{array}{rcl} \displaystyle \int_0^1 \log\big(1 - e^{\sqrt{x}}\big)\,dx & = & \displaystyle \int_0^1 \ln\big(e^{\sqrt{x}} - 1\big)\,dx + \pi i \; = \; \int_0^1 \left\{ \sqrt{x} + \ln\big(1 - e^{-\sqrt{x}}\big)\right\}\,dx + \pi i \\ & = & \displaystyle \tfrac23 + 2\int_0^1 \ln(1 - e^{-u})\,u\,du + \pi i \; =\; \tfrac23 + 2\sum_{n=1}^\infty \tfrac{1}{n}\int_0^1 ue^{-nu}\,du + \pi i \\ & = & \displaystyle \tfrac23 + 2\sum_{n=1}^\infty \left\{ \tfrac{1}{n^2}e^{-n} + \tfrac{1}{n^3}e^{-n} - \tfrac{1}{n^3}\right\} + \pi i \\ & = & 2\left\{ \mathrm{Li}_2(e^{-1}) + \mathrm{Li}_3(e^{-1}) - \zeta(3) + \tfrac13\right\} + \pi i \end{array}$ making the answer $2+2+1+3+1+3+3-2 = \boxed{13}$ .

Change in normal integrals

The answer is 13.

1 solution

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