my friend ask me about this:
If $\phi{(u)}=\ln{\frac{e^{\theta u}-1}{e^{\theta}-1}}$, then prove $1+4\int_{0}^{1}\frac{\phi{(u)}}{\phi'{(u)}}du=1-\frac{4}{\theta}(1-D_1(\theta))$.
where $D_1(\theta)=\frac{1}{\theta}\int_{0}^{\theta}\frac{t}{e^{t}-1}dt$ is Debye Function.
i tried to integrate $\frac{\phi{(u)}}{\phi'{(u)}}$, but it is in fact hard.
How to prove it? any special technic to do it?
I think the suggested formula is incorrect. If it were true, then one would have $$ \int_0^1\frac{\phi(u)}{\phi'(u)}\,du=\frac{1}{\theta}\bigl(D_1(\theta)-1\bigr). $$ Let us work on the integral in the left-hand side. First we replace $u$ by $s/\theta$, and get $$ \int_0^1\frac{\phi(u)}{\phi'(u)}\,du =\frac{1}{\theta^2}\int_0^\theta(1-e^{-s})\ln\Bigl(\frac{e^s-1}{e^{\theta}-1}\Bigr)\,ds. $$ Integrating by parts, moving a derivative to the logarithmic term, you find that this equals $$ \frac{1}{\theta^2}\biggl(\Bigl[(s+e^{-s}-1)\ln\Bigl(\frac{e^s-1}{e^{\theta}-1}\Bigr)\Bigr]_0^{\theta}-\int_0^\theta (s+e^{-s}-1)\frac{e^s}{e^{s}-1}\,ds\biggr). $$ Now, we are lucky. The out-integrated part disappears (we choose the $-1$ for this to happen), and the integrand in the last integral can be simplified to $$ \frac{s}{e^s-1}+s-1. $$ Thus, we end up with $$ \int_0^1\frac{\phi(u)}{\phi'(u)}\,du = -\frac{1}{\theta^2}\int_0^{+\theta}\frac{s}{e^s-1}+s-1\,ds=\frac{1}{\theta}\bigl(1-D_1(\theta)\bigr)-\frac{1}{2}. $$ As you can see, this differ both with the sign and with a constant.