Consider the following integral $$ \int_{-\infty}^{-(-a+\frac{b}{y})^{1/2}}\frac{e^{-x^2}}{x}dx $$ where $a,b>0$. The integral can be solved exactly but I am not interested in that. I want to perform the expansion of the integral around $y=0$, maybe using something analogous to Leibnitz's rule for computing derivatives of integrals. I am aware that the dependence on $y$ in the integration limits is not differentiable around $y=0$ so I expect the result to be something like $$ e^{\frac{c_0}{y}}(c_1+c_2y+c_3y^2+\ldots) $$ but I have no idea how to approach this. So, how can I tackle this?
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$$\int_{-\infty}^{-(-a+\frac{b}{y})^{1/2}}\frac{e^{-x^2}}{x}dx=\tfrac{1}{2}{\rm Ei}\,(a-b/y)=$$ $$\qquad\qquad=\tfrac{1}{2}(a-b/y)^{-1}\exp(a-b/y)\sum_{n=0}^\infty\frac{n!}{(a-b/y)^n}$$ $$\qquad\qquad=-\frac{y }{2 b}e^{a-\frac{b}{y}}\sum_{n=0}^\infty (-1)^{n}n!(1-a)^n(y/b)^n$$
answered May 4, 2017 at 15:27
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$\begingroup$ I just expanded the exponential integral function $\endgroup$ Commented May 4, 2017 at 15:31
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$\begingroup$ Care to elaborate more please? $\endgroup$ Commented May 4, 2017 at 15:34
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$\begingroup$ but how do you get the expansion in the second line? $\endgroup$ Commented May 4, 2017 at 15:39
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