Approximating an indefinite integral via a power series expansion I came across an integral in literature that is approximated in a certain way, but I don't really follow the mathematical justification. In a minimal example the integral reads as, $$\int_0^{\infty} dx \frac{f(x)}{x(y-x)}.$$ Here $y < 0$ and $f(x)$ is some function for which the expansion for small $x$ is known to lowest order as, $$f(x) \approx a \sqrt{x}, \qquad x \ll 1.$$ Here $a$ is some constant. Other than this $f(x)$ is arbitrary, but for $x \rightarrow 
\infty$ it must be well behaved such that the integral is convergent. Now we aim to approximate this integral for small $|y|$, and the way this is done is by saying that to lowest order in $|y|$ this integral may be written as,  $$\int_0^{\infty} dx \frac{f(x)}{x(y-x)} \approx \int_0^{\infty} dx \frac{a\sqrt{x}}{x(y-x)} = -\frac{a\pi}{\sqrt{|y|}} \qquad |y| \ll 1.$$ I don't really know how to justify this. Taking a simple example of $f(x) = \sin(a\sqrt{x})$ I can derive that this holds, but in the general case I am not so sure. What I attempted was a change of variables $x = y z$. In that case,$$\int_0^{\infty} dx \frac{f(x)}{x(y-x)} = \int_0^{\infty} dz \frac{f(yz)}{yz(1-z)}.$$ The integral on the right hand side can be Taylor expanded into powers of $y$, which will indeed give the lowest order term that I am looking for. However higher terms in the expansion will contain diverging integrals, such that it seems wrong to me to just neglect them. Is there perhaps a better way to show that the approximation holds? Or am I missing something?
Thanks!
 A: The basic idea is to note that while the expansion in $f$ diverges in general, it is convergent near zero, even for small $y$'s. On the other hand you are assured than the tail integral converges, so one splits the integral into to ranges $I_1=\int_0^\Lambda$;$I_2=\int_\Lambda^\infty$. The second integral converges for $y=0$ to a $y$ independent constant $I_2(\Lambda)$ while the first integral contains all the $y$ dependent singular behavior.  Higher order terms can be extracted by expanding $I_2$ in $y$ and expanding $I_1$ using the expansion for $f$.
More explicitly:
Split the integral into 2 parts:
$I=I_1+I_2$
$I_1 =\int_0^\Lambda dx{f(x)\over x (y-x)} $, $I_2=\int_\Lambda^{\infty} dx {f(x)\over x (y-x)}$
In the limit of $y \rightarrow 0$, Expand $I_2$ in $y$:
$$I_2=\int_\Lambda^{\infty}dx{f(x)\over x (y-x)}\sim\int_{\Lambda}^\infty -{f(x)\over x^2}+O(y)$$
and $I_1$  using the expansion for $f$
$$I_1=\int_0^\Lambda dx{a\sqrt x +H.O.T \over x (y-x)}$$
While we cannot take the $\Lambda$ limit  in $I_1$ to infinity without seeing a divergence in the non singular terms, we can estimate the first order term $$I_{1;0}=\int_0^\Lambda dx{a\sqrt x\over x (y-x)}$$ by extending to infinity and then subtracting the finite part:
$$I_{1;0}=\int_0^\Lambda dx{a\sqrt x  \over x (y-x)}=\int_0^\infty dx{a\sqrt x  \over x (y-x)} - \int_{\Lambda}^\infty dx{a\sqrt x \over x (y-x)}$$
Expanding the integrals in $y$ is now possible in principle, and the $\Lambda$ terms should cancel term by term in $O(y)$
