I'm trying to evaluate the following integral:
$$\int_{0}^{1}\ln \left( \frac{e^{-\theta t}- 1}{e^{-\theta}-1} \right) \left(1 - e^{\theta t}\right)dt, ~~\theta \neq 0$$
However, I'm having trouble with it, not really sure how to tackle this. Any suggestions on how to proceed would really help!
We have
$$I=\int_0^1\ln\left(\frac{1-e^{-\theta t}}{1-e^{-\theta}}\right)\left(1-e^{\theta t}\right)dt=\int_0^1\ln\left(\frac{1-e^{-\theta t}}{1-e^{-\theta}}\right)dt-\int_0^1\ln\left(\frac{1-e^{-\theta t}}{1-e^{-\theta}}\right)e^{\theta t}dt$$
Splitting the logs and distributing yields
$$I=\int_0^1\ln(1-e^{-\theta t})dt-\int_0^1\ln(1-e^{-\theta})dt-\int_0^1\ln(1-e^{-\theta t})e^{\theta t}dt+\int_0^1\ln(1-e^{-\theta})e^{\theta t}dt$$
The second and fourth integrals are trivial as each integrand is independent of $t$
$$I=\int_0^1\ln(1-e^{-\theta t})dt-\ln(1-e^{-\theta})-\int_0^1\ln(1-e^{-\theta t})e^{\theta t}dt+\frac{\ln(1-e^{-\theta})}{\theta}\left(e^\theta-1\right)$$
The first integral is able to be written using a power series for $\theta>0$
$$\ln(1-e^{-\theta t})=-\sum_{n=1}^\infty\frac{e^{-\theta tn}}{n}$$
So
$$\int_0^1\ln(1-e^{-\theta t})dt=-\int_0^1\sum_{n=1}^\infty\frac{e^{-\theta tn}}{n}dt=\frac{1}{\theta}\sum_{n=1}^\infty\frac{e^{-\theta n}-1}{n^2}=\frac{1}{\theta}\left(\sum_{n=1}^\infty\frac{e^{-\theta n}}{n^2}-\sum_{n=1}^\infty\frac{1}{n^2}\right)$$
The first sum is the definition of the polylog and the other is very well known
$$\boxed{\int_0^1\ln(1-e^{-\theta t})dt=\frac{1}{\theta}\left(\operatorname{Li}_2\left(e^{-\theta}\right)-\frac{\pi^2}{6}\right)}$$
That leaves only one integral remaining - we can try the same trick
$$e^{\theta t}\ln(1-e^{-\theta t})=e^{\theta t}\sum_{n=1}^\infty\frac{-e^{-\theta tn}}{n}=-\sum_{n=1}^\infty\frac{e^{\theta t(1-n)}}{n}$$
So
$$\int_0^1\ln(1-e^{-\theta t})e^{\theta t}dt=-\int_0^1\sum_{n=1}^\infty\frac{e^{\theta t(1-n)}}{n}dt=\frac{1}{\theta}\sum_{n=1}^\infty\frac{e^{-\theta(n-1)}-1}{n(n-1)}$$
The very first term is actually indeterminant so we should resolve that using L'Hôpital's rule
$$\lim_{n\to1}\frac{e^{-\theta(n-1)}-1}{n^2-n}=\lim_{n\to1}\frac{-\theta e^{-\theta(n-1)}}{2n-1}=-\theta$$
Then the integral becomes
$$\int_0^1\ln(1-e^{-\theta t})e^{\theta t}dt=-1+\frac{1}{\theta}\sum_{n=2}^\infty\frac{e^{-\theta(n-1)}-1}{n(n-1)}$$
Where we can split the terms and denominator with partial fractions
$$\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}$$
The first sum down below is what we used to go to infinite series in the first place but with $n=1$ removed. The second is the same except it begins at $n=2$ where $n=1$ is now valid again, so we have to subtract the first term, which is negative, resulting in adding it. The last series is telescoping which is easily evaluated - the only term that doesn't cancel in it is the very first.
$$\boxed{\int_0^1\ln(1-e^{-\theta t})e^{\theta t}dt=-1+\frac{1}{\theta}\left(\underbrace{\sum_{n=2}^\infty\frac{e^{-\theta(n-1)}}{n-1}}_{-\ln(1-e^{-\theta})}+e^\theta\underbrace{\sum_{n=2}^\infty\frac{-e^{-\theta n}}{n}}_{\ln(1-e^{-\theta})+e^{-\theta}}-\underbrace{\sum_{n=2}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)}_{1}\right)}$$
Finally the last integral is known and we have for $\theta>0$
$$I=\frac{1}{\theta}\left(\operatorname{Li}_2\left(e^{-\theta}\right)-\frac{\pi^2}{6}\right)-\ln\left(1-e^{-\theta}\right)+1-\frac{\ln\left(1-e^{-\theta}\right)}{\theta}\left(e^\theta-1\right)+\frac{\ln\left(1-e^{-\theta}\right)}{\theta}\left(e^\theta-1\right)$$
$$\boxed{\therefore\int_0^1\ln\left(\frac{1-e^{-\theta t}}{1-e^{-\theta}}\right)\left(1-e^{\theta t}\right)dt=1+\frac{1}{\theta}\left(\operatorname{Li}_2\left(e^{-\theta}\right)-\frac{\pi^2}{6}\right)-\ln\left(1-e^{-\theta}\right)}$$