I have to show that an operator $A:L_2[0,1]\rightarrow L_2[0,1]$ given by $$Ax(t)=\int_0^1 \frac{\cos(t\tau)}{|t-\tau|^\frac{1}{5}}\ x(\tau)\ d\tau$$ is compact.
My idea is to show that it is uniformly limited and uniformly continuous, but I'm having difficulties checking the 2nd condition.
For it I take $t_1,t_2 \in [0,1]$ and compute $|Ax(t_1)-Ax(t_2)|$ and it goes:
$$|Ax(t_1)-Ax(t_2)|\leq\int_0^1|\frac{|t_2-\tau|^\frac{1}{5}\cos(t_1\tau) - |t_1-\tau|^\frac{1}{5}\cos(t_2\tau)}{|t_1-\tau|^\frac{1}{5}|t_2-\tau|^\frac{1}{5}}||x(\tau)|d\tau$$ $$\leq\text{max}_{\tau\in[0,1]}|x(\tau)|\int_0^1\frac{|t_2-\tau|^\frac{1}{5} + |t_1-\tau|^\frac{1}{5}}{|t_1-\tau|^\frac{1}{5}|t_2-\tau|^\frac{1}{5}}d\tau$$
I used triangular inequality and that $|\cos(t)|\leq1$ on the numerator. The problem is that calculating this integral doesn't give me anything of the form $|t_1-t_2|$.