For what $\alpha > 0$ do we have $\int_{D_{\alpha}} f \ d\lambda < + \infty$ ? Problem: Define for $\alpha > 0$ the subset of $\mathbb{R}^2$ given as $$ D_{\alpha} = \left\{(x,y) \in \mathbb{R}^2 \mid 0 < y, 0 < x < y^{\alpha} < 1 \right\}. $$
Define the function $$f : (0, \infty) \times (0, \infty) \to [0, \infty): (x,y) \mapsto \frac{1}{(x+y)^2}. $$ For which values of $\alpha > 0$ is $$ \int_{D_{\alpha}} f \ d\lambda < + \infty $$ ? 
Attempt: I first wanted to change coordinates to polar coordinates. But I don't know how to redefine the subset $D_{\alpha}$ in terms of polar coordinates. Since $f$ is a positive function, we can switch order of integration (Fubini). So I wrote (I'm not sure if the integration bounds are correct) $$ \int_{D_{\alpha}} f \ d \lambda = \int_{0}^1 \bigg( \int_{0}^{y^{\alpha}} \frac{1}{(x+y)^{2}} dx \bigg) dy = \int_{0}^{y^{\alpha}} \bigg( \int_0^1 \frac{1}{(x+y)^2} dy \bigg) dx = \int_{0}^{y^{\alpha}} \bigg( \frac{1}{x} - \frac{1}{x+1} \bigg) dx. $$ I'm not sure how to proceed. I can only use linearity if I know the two terms inside the brackets are integrable. But $$ \int_{0}^{y^{\alpha}} \frac{dx}{x} = \lim_{\epsilon \to 0} \ln \bigg( \frac{y^{\alpha}}{\epsilon} \bigg) $$ and I don't think I can get this finite for any $\alpha$. 
Help/advice with this problem is appreciated.
 A: $$\int_{D_\alpha}fd\lambda=\int_0^1\int_0^{y^\alpha}\frac{1}{(x+y)^2}dxdy=\int_0^1\left[\frac{-1}{x+y}\right]_0^{y^\alpha}dy=\int_0^1 \frac{1}{y}-\frac{1}{y+y^\alpha}dy=[\ln[y]]_0^1-\left[\frac{\alpha\ln[y]-\ln[y^\alpha+y]}{\alpha-1}\right]_0^1 =\lim_{\epsilon\to0}-\ln[\epsilon]+\frac{\ln[2]}{\alpha-1}+\lim_{\epsilon\to0}\dfrac{\alpha\ln[\epsilon]-\ln[\epsilon]}{\alpha-1}$$
so we want:
$$\dfrac{\alpha\ln[\epsilon]-\ln[\epsilon]}{\alpha-1}-\ln[\epsilon]=0$$
but it is true for any choice of $\alpha$ and then the only restriction is $\alpha \neq 1$
A: The integral equals
$$\tag 1 \int_0^1 \left (\frac{1}{y}-\frac{1}{y+y^\alpha}\right)\,dy.$$
as @Cuoredicervo found. Suppose $0<\alpha <1.$ Note we have
$$\frac{1}{y+y^\alpha} < \frac{1}{y^\alpha}.$$
Because $\int_0^1 (1/y^\alpha)\, dy$ converges, so does $\int_0^1 1/(y+y^\alpha)\, dy.$ Since $\int_0^1 (1/y)\, dy$ diverges, and divergent - convergent = divergent, $(1)$ diverges.
If $\alpha = 1,$ then $(1)= \int_0^1 1/(2y)\, dy = \infty.$
If $\alpha >1,$ then
$$\frac{1}{y}-\frac{1}{y+y^\alpha} = \frac{y^\alpha}{y(y+y^\alpha)}< \frac{y^\alpha}{y^2} = y^{\alpha -2}.$$
Since $\alpha -2 > -1,$ $\int_0^1  y^{\alpha -2}\, dy$ converges.
Conclusion: The integral diverges for $0<\alpha \le 1,$ and converges for $\alpha > 1.$
