Evaluate:$$\int_0^{\ln 2}\left(\frac{1}{e^x-1}-\left\lfloor\frac{1}{e^x-1}\right\rfloor\right)dx$$
My Attempt:
Upon evaluating $$\int_0^{\ln 2}\frac{1}{e^x-1}dx$$ I am getting $$|\ln(e^x-1)-x|_0^{\ln 2}$$
And on evaluating $$\int_0^{\ln 2}\left\lfloor\frac{1}{e^x-1}\right\rfloor dx$$ by substituting $\frac{1}{e^x-1}=t$ i.e. $x=\ln(1+\frac{1}{t})$
I get
$$\int_0^{\ln 2}\left\lfloor\frac{1}{e^x-1}\right\rfloor dx=\int_1^{\infty}\frac{\left\lfloor t\right\rfloor}{t(1+t)}dt=\lim_{n\to \infty}\left(\ln(n+1)-n\ln(1+\frac{1}{n})\right)$$
Given answer is $1-\ln 2$
which I will get if I cancel out the infinities I obtain in solving $\int_0^{\ln 2}\frac{1}{e^x-1}dx$
and $\int_0^{\ln 2}\left\lfloor\frac{1}{e^x-1}\right\rfloor dx$.
But it is not justified as limit of the form $\infty-\infty$ is indeterminate.
So I think solving the two integrals separately may not be a good idea.
Can the integral be evaluated as a whole.
Best Answer
Following your idea, we perform the change of integration variables from $x$ to $t$ via $t=1/(\mathrm{e}^x-1)$. Then \begin{align*} \int_1^{ + \infty } {\frac{{t - \left\lfloor t \right\rfloor }}{{t(t + 1)}}{\rm d}t} &= \sum\limits_{k = 1}^\infty {\int_k^{k + 1} {\frac{{t - \left\lfloor t \right\rfloor }}{{t(t + 1)}}{\rm d}t} } = \sum\limits_{k = 1}^\infty {\int_0^1 {\frac{{t - \left\lfloor t \right\rfloor }}{{(t + k)(t + k + 1)}}{\rm d}t} }\\ & = \int_0^1 {(t - \left\lfloor t \right\rfloor )\sum\limits_{k = 1}^\infty {\frac{1}{{(t + k)(t + k + 1)}}} {\rm d}t} \\& = \int_0^1 {(t - \left\lfloor t \right\rfloor )\sum\limits_{k = 1}^\infty {\left( {\frac{1}{{t + k}} - \frac{1}{{t + k + 1}}} \right)} {\rm d}t} \\ & = \int_0^1 {\frac{{t - \left\lfloor t \right\rfloor }}{{t + 1}}{\rm d}t} = \int_0^1 {\frac{{t}}{{t + 1}}{\rm d}t} = 1 - \log 2. \end{align*}