Prove $\int_0^B e^\left(x^2\right) \;dx \sim \frac{e^\left(B^2\right)}{2B}$ where $B \to \infty$ Prove that:
$$\int_0^B e^\left(x^2\right) \;dx \sim \frac{e^\left(B^2\right)}{2B}$$ where $$B \to \infty$$
We should prove the equivalence of the given integral to the expression in the right part. 
It's obvious, that the integral in the given form cannot be expressed through known indefinite integrals. So let's try to apply a method of replacing a variable. 
Write 
$$
 \begin{matrix}
  u = x^2 \\
  dx = \frac{1}{2} u^\left(-\frac{1}{2}\right) du \\
        \end{matrix}
$$
Now our integral looks like: 
$$\frac{1}{2}\int_0^B e^\left(u\right) \sqrt{u} \;du$$
This is kinda cryptic. If we ask Wolfram to calculate indefinite integral for this integrand, we'll get: 
$$\frac{1}{2}\int e^\left(u\right) \sqrt{u} \;du = e^u \sqrt{u} - \frac{1}{2} erfi(\sqrt{u}) + C$$
It's been calculated with a help of erfi function.
I'm sure this way isn't close to the solution at all. So please give me at least a hint of how this integral can be presented in an equivalent form...
 A: Using the same approach as in this answer, but leaving out the details, which are handled in a similar fashion, we get the asymptotic expansion
$$
\begin{align}
\int_0^xe^{t^2}\,\mathrm{d}t
&=e^{x^2}\int_0^xe^{t^2-x^2}\,\mathrm{d}t\\
&=e^{x^2}\int_0^xe^{-2tx+t^2}\,\mathrm{d}t\\
&=xe^{x^2}\int_0^1e^{-(2t-t^2)x^2}\,\mathrm{d}t\\
&\sim xe^{x^2}\int_0^\infty e^{-2ux^2}\left(1+u+\frac32u^2+\frac52u^3+\frac{35}8u^4+O\left(u^5\right)\right)\,\mathrm{d}u\\
&=\left(\frac1{2x}+\frac1{4x^3}+\frac3{8x^5}+\frac{15}{16x^7}+\frac{105}{32x^9}+O\left(\frac1{x^{11}}\right)\right)e^{x^2}
\end{align}
$$
where $2u=2t-t^2$.
A: Integrate by parts:
$$\begin{align}\int_0^B dx \, e^{x^2} &= \underbrace{\int_0^1 dx \, e^{x^2}}_{C} + \int_1^B dx \, e^{x^2}\\ &= C + \int_1^B dx \,x \frac{e^{x^2}}{x} \\ &= C + \frac12 \left [\frac{e^{x^2}}{x} \right ]_1^B + \frac12 \int_1^B dx \, \frac{e^{x^2}}{x^2}\\ &=  \underbrace{C - \frac{e}{2}}_{\text{constant which is dominated by other terms}} +  e^{B^2} \left [\frac1{2 B} + O \left ( \frac1{B^3}\right ) \right ]  \end{align}$$
Thus, as $B \to \infty$, the integral behaves as
$$\int_0^B dx \, e^{x^2} =  e^{B^2} \left [\frac1{2 B} + O \left ( \frac1{B^3}\right ) \right ] $$
