# Definite integral of $|t-s|^{2a-2}$

Hi I'm looking for some help with the function $$f(s,t)=a(2a-1)|t-s|^{2a-2}.$$ I'm trying to solve the following definite integral $$\int_0^t\int_0^sf(u,v)dudv=a(2a-1)\int_0^t\int_0^s|u-v|^{2a-2}dudv.$$

I know that we have $$\int_0^t\int_0^sf(u,v)dudv=\frac{1}{2}\left(s^{2a}+t^{2a}-|s-t|^{2a}\right).$$ I just don't know how to get there.

• What are you trying to solve precisely? – Szeto Jun 23 '18 at 8:26

The formula is correct if $t \geq 0, s \geq 0$.
You can consider the cases $t<s$ and $t>s$ separately and split the integration range into subranges where $u<v$ and $u>v$ for each case.
Alternatively, you can write $$a (2a-1) \int_0^t \int_0^s ((u-v)^2)^{a-1} du dv = \\ a \int_0^t (u-v) ((u-v)^2)^{a-1} \Big\rvert_{u=0}^s dv = \\ a \int_0^t (s-v) ((s-v)^2)^{a-1} dv + a \int_0^t v (v^2)^{a-1} dv = \\ -\frac 1 2 ((s-v)^2)^a \bigg\rvert_{v = 0}^t + \frac 1 2 (v^2)^a \bigg\rvert_{v = 0}^t = \\ \frac 1 2 (|s|^{2a} + |t|^{2a} - |s-t|^{2a}).$$ This is valid for any real $t,s$.