integration - Find values so that integral is a bounded operator

Find all positive values $\alpha$ for which the formula

$$\begin{align*} A_\alpha u(x) = \int\limits_0^1 \frac{u(y)}{(x+y)^\alpha} \,dy \end{align*}$$
defines a bounded operator in $L^1([0,1])$. Compute its norm.

I know that this defines a bounded operator for $\alpha <1$. What I did will not work for $\alpha \geq 1$. I tired to use specific $L^1$ functions but its not working. Do you have any advice on how to approach these types of problems?

$$\begin{align*} ||A_\alpha u(x)||_1 & = \int\limits_0^1\left| \int\limits_0^1 \frac{u(y)}{(x+y)^{\alpha}}\, dy \right|\,dx \\ & \leq \int\limits_0^1 \int\limits_0^1 \frac{|u(y)|}{(x+y)^{\alpha}}\,dy\, dx \\ & \leq \int\limits_0^1 \int\limits_0^1 \frac{|u(y)|}{x^{\alpha}}\,dy\, dx\\ & =\int\limits_0^1 ||u||_1 \frac{1}{x^{\alpha}}\, dx\\ \end{align*}$$

The integral above converges if $\alpha <1$.