$\lim_{x\to \infty} \left(\int_0^x e^{t^2} dt\right)^{\frac 1 {x^2}}$

Question

Prove that $$e=\lim_{x\to \infty} \left(\int_0^x e^{t^2} \mathrm{d}t\right)^{\frac {1}{x^2}}.$$

I tried to take the logarithm of this expression and use the integral mean value theorem - for all $x>0$ there exists a point $c\in (0,x)$ such that $e^{c^2} =\frac 1 x \int_0^x e^{t^2} \mathrm{d}t$, so by taking logarithm we have $$\ln \left( \left( \int_0^x e^{t^2} \mathrm{d} t \right)^{\frac{1}{x^2}}\right)=\frac{1}{x^2} \ln (x\cdot e^{c^2})=\frac{c^2}{x^2}\ln x.$$ If we establish a better bound on $c$ this might be useful; but I couldn't find a continuation.

Answer 1

A proof without using L'Hospital's rule.

Since $\;e^{2t}\geqslant2t+1>2t-2\;\;$ for any $\,t\in(-\infty,+\infty)\;,$

it follows that

$(2t-2)e^{t^2-2t}<e^{t^2}\quad$ for any $\,t\in(-\infty,+\infty)\;\;,$

$\displaystyle\int_0^x(2t-2)e^{t^2-2t}\,\mathrm dt<\int_0^x e^{t^2}\mathrm dt\quad$ for any $\,x\in(0,+\infty)\;\;,$

$\displaystyle\int_0^x e^{t^2-2t}\,\mathrm d\!\left(t^2-2t\right)<\int_0^x e^{t^2}\mathrm dt\quad$ for any $\,x\in(0,+\infty)\;\;,$

$\displaystyle e^{x^2-2x}-1=\left[e^{t^2-2t}\right]_0^x<\int_0^x e^{t^2}\mathrm dt\quad$ for any $\,x\in(0,+\infty)\;.$

On the other hand , for any $\,x\in[3,+\infty)\,,$

$\displaystyle e^{x^2-2x-1}<2e^{x^2-2x-1}-1=\dfrac2ee^{x^2-2x}-1<e^{x^2-2x}-1\;.$

Consequently ,

$\displaystyle e^{x^2-2x-1}<\!\int_0^x\!e^{t^2}\mathrm dt<xe^{x^2}\!\!<e^xe^{x^2}\!\!=e^{x^2+x}\quad\forall\,x\!\in[3,+\infty)\;,$

$\displaystyle\left(e^{x^2-2x-1}\right)^{\frac1{x^2}}\!\!<\left(\!\int_0^x\!e^{t^2}\mathrm dt\!\right)^{\!\frac1{x^2}}\!\!<\left(e^{x^2+x}\right)^{\frac1{x^2}}\quad\forall\,x\!\in[3,+\infty)\;,$

$\displaystyle e^{1-\frac2x-\frac1{x^2}}<\left(\int_0^x\!e^{t^2}\mathrm dt\right)^{\!\frac1{x^2}}\!<e^{1+\frac1x}\quad\forall\,x\!\in[3,+\infty)\;.\quad\color{blue}{(*)}$

Since $\,\lim\limits_{x\to+\infty}e^{1-\frac2x-\frac1{x^2}}=\lim\limits_{x\to+\infty}e^{1+\frac1x}=e\;,\;$ from $\,(*)\,$ by applying the Squeeze Theorem, it follows that

$\displaystyle\lim\limits_{x\to+\infty} \left(\int_0^x\!e^{t^2}\mathrm dt\right)^{\!\frac1{x^2}}=e\;.$

Answer 2

An integration by parts yields: $$ \int_{1}^{x} \mathrm{e}^{t^2}\ \mathrm{d}t = \int_{1}^{x} \frac{1}{2t}\,2t\,\mathrm{e}^{t^2}\ \mathrm{d}t = \frac{\mathrm{e}^{x^2}}{2x}-\frac{\mathrm{e}}{2}+\frac{1}{2}\int_{1}^{x} \frac{\mathrm{e}^{t^2}}{t^2}\ \mathrm{d}t. $$ Since $\frac{\mathrm{e}^{t^2}}{t^2}=\mathrm{o}(\mathrm{e}^{t^2})$ as $t$ tends to $+\infty$ and since $t\mapsto \mathrm{e}^{t^2}$ is positive and not integrable over $[1,{+\infty}[$, we have: $$ \int_{1}^{x} \frac{\mathrm{e}^{t^2}}{t^2}\ \mathrm{d}t=\mathrm{o}\left(\int_{1}^{x} \mathrm{e}^{t^2}\ \mathrm{d}t\right). $$ This implies that: $$ \int_{1}^{x} \mathrm{e}^{t^2}\ \mathrm{d}t\underset{x\to{+\infty}}{\sim} \frac{\mathrm{e}^{x^2}}{2x} \qquad \text{and then}\qquad \int_{0}^{x} \mathrm{e}^{t^2}\ \mathrm{d}t\underset{x\to{+\infty}}{\sim} \frac{\mathrm{e}^{x^2}}{2x}. $$ Consequently, we deduce that: $$ \ln\left(\int_{0}^{x} \mathrm{e}^{t^2}\ \mathrm{d}t\right)\underset{x\to{+\infty}}{\sim} x^2 $$ and then: $$ \frac{1}{x^2}\ln\left(\int_{0}^{x} \mathrm{e}^{t^2}\ \mathrm{d}t\right)\underset{x\to{+\infty}}{\sim} 1. $$

Finally, by the continuity of the exponential function, we obtain:

$$ \lim_{x\to{+\infty}} \left(\int_{0}^{x} \mathrm{e}^{t^2}\ \mathrm{d}t\right)^{1/x^2}=\mathrm{e} $$