Find $ \lim_{x\to\infty} x \int_{0}^{x} e^{t^2-x^2}\, dt$

Question

I'm stuck on part b) of this problem from the 1989 edition of Elements of Calculus and Analytic Geometry by Thomas and Finney, Chapter 12 (Power Series), Section 6 (Indeterminate Forms). Part a) asks to prove $$ \lim_{x\to\infty}\int_{0}^{x} e^{t^2}\, dt$$ which is straightforward. I substituted a power series, integrated a few terms, and saw that it $\rightarrow \infty$.
Part b) asks to find $$ \lim_{x\to\infty} x \int_{0}^{x} e^{t^2-x^2}\, dt$$ I've tried it a couple ways, first rewriting the integrand as $\frac{e^{t^2}}{e^{x^2}}$. I then treated the denominator as a constant and brought it in front of the integral with the $x$. I then integrated a power series for the remaining integral; I keep getting an answer of $\infty$, but the given answer is $\frac{1}{2}$.

Alternatively, I tried substituting $(t^2-x^2)$ into a generic $e^x$ power series before integrating, but that also gives me $\infty$. (I'm too slow with LaTeX to show all my work here.) I can't find my mistake, and I can't think of what else to try.

Is my technique faulty or is there some methodological hint?

Answer 1

You can use l'Hospital, $$ \lim_{x\to+\infty}\frac{\int_0^x e^{t^2}\,dt}{e^{x^2}/x}=\lim_{x\to+\infty}\frac{e^{x^2}}{2e^{x^2}-e^{x^2}/x^2}=\lim_{x\to+\infty}\frac{1}{2-1/x^2}=\frac{1}{2}. $$

Answer 2

If you wish to avoid the usage of L'hospital rule using simple estimates you may proceed as follows:

For a lower bound \begin{aligned} x e^{-x^2}\int^x_0 e^{t^2}\,dt \geq e^{-x^2}\int^x_0 t e^{t^2}\,dt=\frac12(1-e^{-x^2})\xrightarrow{x\rightarrow\infty}\frac12 \end{aligned}

For an upper bound

\begin{aligned} x e^{-x^2}\int^x_0 e^{t^2}\,dt &=x\int^x_0 e^{(t-x)(t+x)}\,dt\leq x\int^x_0 e^{(t-x)2x}\,dt\\ &=x e^{-2x^2}\int^x_0e^{2xt}\,dt= \frac{1-e^{-2x^2}}{2}\xrightarrow{x\rightarrow\infty}\frac12 \end{aligned}

The advantage of not using L'Hospital rule here is that you also obtain rate of convergence. The function $F(x)=x e^{-x^2}\int^x_0 e^{t^2}\,dt$ converges to$\frac12$ much faster than $G(x)=\frac{1}{2-x^{-2}}$ does.