I want to evaluate the following integral:
$$I(n)=\int_0^1 t^n \log(t^4-t^2+1) dt$$
The motivation is that, for odd values of $n$, Wolfram suggests beautiful closed forms, but I can't see the pattern. For even values of $n$ it does find closed forms, but not always.
Therefore I want to see if there is a general closed form for the general case, that could explain this even-odd disparity.
By integrating by parts, I got the following integral:
$$I(n)=-\frac{1}{n+1}\int_0^1\frac{t^{n+1}(4t^3-2t)}{t^4-t^2+1}dt$$
From here, doing partial fractions with the four roots of the denominator, $+e^{i\frac{\pi}{6}}$, $-e^{i\frac{\pi}{6}}$, $+e^{-i\frac{\pi}{6}}$, $-e^{-i\frac{\pi}{6}}$, we get four integrals of the form $$\int_0^1\frac{t^{n+1}(4t^3-2t)}{t+a}dt$$
But from here? How to proceed?
Any help would be appreciated.
Best Answer
For the odd case $n=2m+1$ \begin{align} I_{2m+1}&=\int_0^1 t^{2m+1} \log(t^4-t^2+1) \overset{t^2\to t}{dt}\\ &= \frac12\int_0^1 t^{m} \log(t^2-t+1) {dt}\\ &= \Re \int_0^1 t^{m} \log(t-a) {dt}\>\>\>\>\>(a=e^{i\frac\pi 3})\\ &\overset{ibp}=-\frac1{m+1}\Re\int_0^1 \frac{t^{m+1}}{t-a}dt\\ &= -\frac1{m+1}\Re \int_0^1 \sum_{k=0}^{m+1}\binom{m+1}k a^k(t-a)^{m-k}\ dt\\ &= -\frac1{m+1}\Re \bigg[\sum_{k=0}^{m}\binom{m+1}k \frac{a^k (t-a)^{m-k+1}}{m-k+1}+a^k\ln(t-a)\bigg]_0^1\\ &=-\frac2{m+1}\sum_{k=0}^m \frac{(-1)^{m-k} \binom{m+1}k}{m-k+1}\sin\frac{\pi(3m+3-2k)}{6}\sin\frac{\pi(m-k+1)}6\\ & \>\>\>\>\> \>\> +\frac\pi{3(m+1)}\sin\frac{\pi(m+1)}3 \end{align} which exhibits the pattern $I_{2m+1}= a+b\cdot\frac\pi{\sqrt3}$, with rational numbers $a$ and $b$. In particular \begin{align} &I_1=\frac\pi{2\sqrt3}-1,\>\>\> I_3=\frac\pi{4\sqrt3}-\frac12,\>\>\> I_5=-\frac1{36}\\ &I_7=\frac5{24}-\frac\pi{8\sqrt3},\>\>\> I_9=\frac{101}{600}-\frac\pi{10\sqrt3},\>\>\> I_{11}=-\frac7{720} \end{align}