How to solve this integral with absolute value

Question

How to solve this integral $$\int_{-a}^{a}(a^t-\lvert{x}\rvert^t)^2dx$$ where $$\lvert{x}\rvert<a$$ and $$t>0$$ I tried distributing the squared and then taking the individual integrals, resulting in $2a^{2t+1}$ but I am not confident in my answer. Any help is appreciated. Thanks.

Answer 1

$$\int_{-a}^{a}(a^t-\lvert{x}\rvert^t)^2dx=\int_0^{a}(a^t-{x}^t)^2dx+\int_{-a}^{0}(a^t-(-x)^t)^2dx.$$

Now let $y=-x$ in the second integral:

$$\begin{aligned}\int_{-a}^{a}(a^t-\lvert{x}\rvert^t)^2dx&=\int_0^{a}(a^t-{x}^t)^2dx+\int_{a}^{0}(a^t-y^t)^2(-dy)\\&=2\int_{0}^{a}(a^t-{x}^t)^2dx\end{aligned}$$

We could have initially observed that the integrand is even, and thus equal to twice the integral from $0$ to $a$, as we found.

Answer 2

It depends if $t$ is an integer or not

$$(a^t-|x|^t)^2=a^{2t}-2a^t|x|^t+|x|^{2t}$$ we can break this integral up into these three parts to simplify (symmetry is useful with $|x|$): $$I_1=\int_{-a}^aa^{2t}dx=a^{2t}(2a)=2a^{2t+1}$$ $$I_2=-2a^t\int_{-a}^a|x|^tdx=-4a^t\int_0^ax^tdx=-4a^t\left[\frac{x^{t+1}}{t+1}\right]_0^a=-\frac{4a^{2t+1}}{t+1}$$ $$I_3=\int_{-a}^a|x|^{2t}dx=2\int_0^ax^{2t}dx=2\left[\frac{x^{2t+1}}{2t+1}\right]_0^a=\frac{2a^{2t+1}}{2t+1}$$ so all together we have: $$I=2a^{2t+1}\left(1-\frac{2}{t+1}+\frac{1}{2t+1}\right)$$ and you should be able to finish this easily from here :)