How do I evaluate this definite integral which blows up at lower limit? I have an integral of the form
 $$\int^{\infty}_{0}{\frac{2a^2-x^{2} }{a^{2}+x^{2}}e^{\frac{-x^{2}}{b^2}}xdx}.$$
On substitution of $x^2=t$ and simplifying, I get integral of the form $$\int^{\infty}_{0}{\frac{e^{-t}}{t}dt}$$ which blows up as $ t \to 0$. Is there any way to approximate it? Is there some cutoff at the lower limit I can use?
 A: That integral is a well known special function, the Exponential Integral . You didn't do the substitution right, though.
If $b$ is $1$, the integral is $\int_0^\infty \frac{a^2-x^2}{a^2+x^2} e^{-x^2} x dx = \int_0^\infty \frac{a^2 - t}{a^2+t} e^{-t} dt$. Then, you $u$-substitute $u=a^2+t$ and shift the bounds from $u=a^2$ to $u=\infty$ and get the integral $\int_{a^2}^\infty \frac{a^2 - (u - a^2)}{u} e^{-u} e^{a^2} du = e^{a^2} (2 a^2 \int_{a^2}^\infty \frac{e^{-u}}{u} du - \int_{a^2}^\infty e^{-u} du)$.  The second integral is simply $1- e^{-a^2}$, and the first integral is the exponential integral evaluated at $a^2$, which is finite for $a \neq 0$.
You can adjust to the case where $b \neq 1$ easily, by factoring out a $b^2$ from the numerator and denominator of the fraction and then substituting $t=\frac{x^2}{b^2}$, and then the rest is similar.
A: Batman gave you a good answer underlining that you missed the lower bound of the integral.
Back to your original notations and taking into account all suggestions made by Batman, for the antiderivative you would arrive to 
$$\int{\frac{2a^2-x^{2} }{a^{2}+x^{2}}e^{\frac{-x^{2}}{b^2}}xdx}=\frac{3}{2} a^2 e^{\frac{a^2}{b^2}}
   \text{Ei}\left(-\frac{a^2+x^2}{b^2}\right)+\frac{1}{2} b^2 e^{-\frac{x^2}{b^2}}$$ and then, using the bounds, the expression of the integral is 
$$\int^{\infty}_{0}{\frac{2a^2-x^{2} }{a^{2}+x^{2}}e^{\frac{-x^{2}}{b^2}}xdx}=\frac{3}{2} a^2 e^{\frac{a^2}{b^2}} \Gamma
   \left(0,\frac{a^2}{b^2}\right)-\frac{b^2}{2}$$
