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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...

mozharovsky
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    Why is there still an $x$ in the evaluated integral? Or does the $\sim$ mean something else? – Irregular User Mar 13 '16 at 20:27
  • sure you don't mean $e^{-x^2}$? – jim Mar 13 '16 at 20:50
  • Despite the poor statement of the question, I answered it because I think there is use to the community. That said, the OP really needs to put some thought into this and other questions. – Ron Gordon Mar 13 '16 at 20:52
  • Thank you all for replying! Sorry for the lack of information. I've updated the first post. And no, it's not $e^\left(-x^2\right)$. – mozharovsky Mar 13 '16 at 20:59
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    The statement can also be proven using L'Hopital: $\lim_{B\to\infty}\frac{\int_0^B e^{x^2}{\rm d}x}{\frac{e^{B^2}}{2B}} = \lim_{B\to\infty}\frac{e^{B^2}}{e^{B^2} - \frac{e^{B^2}}{2B^2}} = 1$. – Winther Mar 13 '16 at 21:07

2 Answers2

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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 $2u=2t-t^2$.

Community
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robjohn
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  • +1: Yep, you get the same expansion by integrating by parts. – Ron Gordon Mar 14 '16 at 01:38
  • As a check, I gave Series[Integrate[Exp[t^2],{t,0,x}],{x,Infinity,10}] to Mathematica 8 (I've not tried Mathematica 10). It returned what I have above, but added a constant $-i\frac{\sqrt\pi}2$ to the answer. While it is true that any constant term is $O\left(\frac1{x^{11}}\right)e^{x^2}$, the question is why would an imaginary constant be added? I assume it is a leftover from the complex method it uses to compute the asymptotic expansion. – robjohn Mar 14 '16 at 14:22
  • absolutely. $Erfi[z] = -i Erf[i z]$ in Mathematica-speak, so no surprise when its Taylor-about-infinity expansion algorithm trips over itself. – Ron Gordon Mar 14 '16 at 14:30
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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

$$\int_0^B dx \, e^{x^2} = e^{B^2} \left [\frac1{2 B} + O \left ( \frac1{B^3}\right ) \right ] $$

Ron Gordon
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