Computing $\int_0^1 \frac{(1-t)(1+at)}{|1+at|^p} dt$.

Question

Just as the title suggests, I want to compute the integral of the form $$ \int_0^1 \frac{(1-t)(1+at)}{|1+at|^p} dt $$ in terms of $a$ and $p$, where $a \in \Bbb R$ and $p\in (1,2)$.

This is not a homework problem or anything, the expression came up when I was trying to calculate something for fun. It doesn't look too hard but the plausible solution I have in my head is a bit long so I decided to post it here to see if there is a faster method.

PS. I'm especially interested in the case $a \le -1$.

Answer 1

For $a\le -1$

\begin{align} & \int_0^1 \frac{(1-t)(1+at)}{|1+at|^p} dt\\ = & \int_0^{-1/a}\underset{y=1+at}{(1-t)(1+at)^{1-p}}dt - \int_{-1/a}^1 \underset{y=-1-at}{ (1-t)(-1-at)^{1-p}}dt \\ = & \frac1{a^2}\int_0^{1}(y-1-a)y^{1-p}dy + \frac1{a^2}\int_0^{-1-a}(y+1+a)y^{1-p}dy\\ =&\frac{ap-3a-1-(-1-a)^{3-p}}{a^2(3-p)(2-p)} \end{align}

Answer 2

Without absolute values we get the indefinite integral equal to

$$\frac{(a (p (t-1)-2 t+3)+1) (a t+1)^{2-p}}{a^2 (p-3) (p-2)}.$$

Now just break up the integral into the parts between $0$ and $-\frac1a$ and $-\frac1a$ to $1.$