# 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.

\(i=\sqrt{-1}\)