Approximating an integral involving a decaying exponential. I've got this integral in a textbook $\Delta N=\int_{E_{0}}^{\infty} \sqrt{E}(e)^{-\frac{E}{k T}} \cdot d E$with
$E_{0}>>K T$.
The author approximates it by saying

The principal contribution to the value of the integral is provided by the smallest values of $E$, namely $E \approx E_0 $ .
The slowly varying factor $\sqrt{E}$ can be taken from under the integral sign if ascribed the constant value $\sqrt{E_{0}}$.

And then proceeds to integrate

$\sqrt{E_{0}} \int_{E_{0}}^{\infty}(e)^{-E / k T} d E$

Can anyone please explain it in a simple way, how can we do the above approximation and not incur massive error.?
 A: The antiderivative for the "true" integral is not very difficult (start with $E=x^2$ and a couple of integrations by parts).
In short, the result is
$$I=\int_{E_0}^\infty \sqrt{E}\, e^{-\frac{E}{k T}}\,dE=\frac{ \sqrt{\pi }}{2} (k T)^{3/2} \text{erfc}\left(\sqrt{\frac{E_0}{k
   T}}\right)+\sqrt{E_0} k T e^{-\frac{E_0}{k T}}$$ while
$$J=\sqrt{E_0}\,\int_{E_0}^\infty  e^{-\frac{E}{k T}}\,dE=\sqrt{E_0}\,k T e^{-\frac{E_0}{k T}}$$
$$\Delta=J-I=\frac{ \sqrt{\pi }}{2} (k T)^{3/2} \text{erfc}\left(\sqrt{\frac{E_0}{k
   T}}\right)$$
Let $E_0= a k T$
$$\Delta=\frac{ \sqrt{\pi }}{2}   (k T)^{3/2}\text{erfc}\left(\sqrt{a}\right)$$ When $a$ is large
$$\text{erfc}\left(\sqrt{a}\right)=\frac{e^{-a}}{\sqrt{\pi a}}\left(1-\frac{1}{2 a}+O\left(\frac{1}{a^2}\right)\right)$$
$$\Delta \sim \frac{(kT)^2}{2\sqrt{E_0}}e^{-\frac{E_0}{k T}}= {E_0}^{3/2}\,\frac{e^{-a}}{2 a^2}$$
$$\left(
\begin{array}{cc}
 a & \frac{e^{-a}}{2 a^2} \\
 3 & 0.002766 \\
 4 & 0.000572 \\
 5 & 0.000135 \\
 6 & 0.000034
\end{array}
\right)$$
