# Integral involving logarithmics and powers: $\int_{0}^{D} z \cdot (\sqrt{1+z^{a}})^{b} \cdot \ln(\sqrt{1+z^{a}})\; \mathrm dz$

Do you guys have any ideas on solving the following integral: $$\int_{0}^{D} z \cdot (\sqrt{1+z^{a}})^{b} \cdot \ln(\sqrt{1+z^{a}})\; \mathrm dz$$ with $a, b$ being constants.

I was trying to use the formula (4.253.1) (shown below) of the book Table of Integrals, Series, and Products, 8th edition, to solve the integral by applying change of random variable, but it seems not to work.

Concerning the antiderivative $$I=\int z \, (\sqrt{1+z^{a}})^{b} \, \ln(\sqrt{1+z^{a}})\, dz$$ let $$\sqrt{1+z^{a}}=t \implies z=\left(t^2-1\right)^{\frac{1}{a}}\implies dz=\frac{2 t \left(t^2-1\right)^{\frac{1}{a}-1}}{a}dt$$ which makes $$I=\frac 2a \int \left(t^2-1\right)^{\frac{2}{a}-1} t^{b+1} \log (t)\,dt$$ and then $$I=\frac{2 }{a \,(2+b)^2}\left(1-t^2\right)^{-2/a} \left(t^2-1\right)^{2/a} t^{b+2}\,J$$where $$J= \, _3F_2\left(1-\frac{2}{a},\frac{b}{2}+1,\frac{b}{2}+1;\frac{b}{2}+2,\frac{b}{2}+2 ;t^2\right)-$$ $$(b+2) \log (t) \, _2F_1\left(1-\frac{2}{a},\frac{b}{2}+1;\frac{b}{2}+2;t^2\right)$$
When $z\to 0$, $t \to 1$ and $$\lim_{t\to 1} \, I=-\frac{(-1)^{-2/a} \Gamma \left(\frac{2}{a}\right) \Gamma \left(\frac{b}{2}+2\right) \left(\psi ^{(0)}\left(\frac{b}{2}+1\right)-\psi ^{(0)}\left(\frac{b}{2}+\frac{2}{a}+1\right)\right)}{a (b+2) \Gamma \left(\frac{b}{2}+\frac{2}{a}+1\right)}$$
• Hi, @Claude, thanks a lot for the answer, just to be sure, then the definite integral with interval from 0 to D can be solved by putting $t=\sqrt{1+D^a}$ into $I$ with hypergeometric functions and then minus $\lim\limits_{t \rightarrow 1} I$, right? – Vic Jan 19 '18 at 9:28
• Great and thanks, @Claude, one more question, I did not follow the step on obtaining the hypergeometric function from the integral involving $t$, do you have any sources or references for that. – Vic Jan 19 '18 at 9:34
• Hi, @Claude, yes, I know about hypergeometric functions, I was just curious about the reference on the equality $\int \left(t^2-1\right)^{\frac{2}{a}-1} t^{b+1} \log (t)\,dt = \frac{1 }{(2+b)^2}\left(1-t^2\right)^{-2/a} \left(t^2-1\right)^{2/a} t^{b+2}\,\cdot \left[ 3F_2\left(1-\frac{2}{a},\frac{b}{2}+1,\frac{b}{2}+1;\frac{b}{2}+2,\frac{b}{2}+2 ;t^2\right)- (b+2) \log (t) \, _2F_1\left(\frac{a-2}{a},\frac{b+2}{2};\frac{b+4}{2};t^2\right) \right]$, which seems quite complicated to me, and you must be very good at hypergeometric functions to find this equality somewhere. Thanks again. – Vic Jan 19 '18 at 9:49