Limit of improper integral $ \lim_{x\to \infty} \ \int_1^x x\,e^{t^2-x^2} \,dt$ I need to calculate the limit of the following improper integral:
$$ \lim_{x\to \infty} \ \int_1^x x\,e^{t^2-x^2} \,dt$$
My solution is $\infty$, but when I enter it in WolframAlpha the solution is $\frac{1}{2}$, so I guess mine is wrong, but I can't figure out where. This is my current solution:
$$
First\ calculate\ the\ integral\ (leaving\ away\ the\ limit\ for\ now)\\[10pt]
Let \
u = t^2-x^2 \\
\frac{du}{dt} = 2t \leftrightarrow dt = \frac{du}{2t} \\[20pt]
\
Substitution:\\[5pt]
\begin{align*}
x \ \int_{1-x^2}^0 x\,e^{u} \frac{du}{2t}
&= \frac{x}{2t} \ \int_{1-x^2}^0 e^{u} \,du \\
&= \frac{x}{2t} \ {\bigl (}{e^u{\bigl )}{\bigl \vert }\,}_{1-x^2}^{0} \\
&= \frac{x}{2t} \ {\bigl (}{e^0-e^{1-x^2}{\bigl )}\,} \\
&= \frac{x}{2t} - \frac{e^{1-x^2} \ x}{2t} \\
&= \frac{x - e^{1-x^2} \ x}{2t} \\
&= \frac{x \ (1-e^{1-x^2})}{2t}
\end{align*}\\
\
\\[20pt]
Now\ calculate\ the\ limit:\\[10pt]
\lim_{x\to \infty} \ \frac{x \ (1-e^{1-x^2})}{2t}
= \infty
$$
What have I done wrong, that I don't get the solution $\frac{1}{2}$
 A: When you do the substitution you cannot put the $1/t$ outside the integral. You should transform it into something with $u$ inside the integral.
Here is another way: The limit you want to calculate, is also the limit of the quotient
$$
\frac{\int_1^x e^{t^2}\,dt}{e^{x^2}/x}.
$$
Differentiating numerator and denominator separately, we get
$$
\frac{\frac{d}{dx}\int_1^x e^{t^2}\,dt}{\frac{d}{dx}e^{x^2}/x}=
\cdots=
\frac{1}{2-1/x^2}.
$$
The limit of the last expression, as $x\to+\infty$, is $1/2$. Thus, by the l'Hospital rule, this is also true for your original problem.
A: It is
$$
\int \limits_1^x x e^{t^2-x^2} dt = \frac{x}{e^{x^2}} \int \limits_1^x e^{t^2} dt = \frac{x}{e^{x^2}} \cdot (G(x) - G(1))
$$
for some differentiable function $G: [1, \infty) \rightarrow \mathbb{R}$ by the fundamental theorem of calculus. Furthermore, $G'(x) = e^{x^2}$, hence $G$ is strictly increasing and $\lim \limits_{x \to \infty} G(x) = + \infty$. Therefore, we are to calculate the limit
$$
\lim \limits_{x \to \infty} \frac{x \cdot G(x)}{e^{x^2}}
$$
This limit is of the form $\frac{\infty}{\infty}$, hence by L'Hospital's rule:
$$
\lim \limits_{x \to \infty} \frac{x \cdot G(x)}{e^{x^2}} \overset{L'H}{=} \lim \limits_{x \to \infty} \frac{x \cdot G'(x) + G(x)}{2x \cdot e^{x^2}} = \frac{1}{2} + \lim \limits_{x \to \infty} \frac{G(x)}{2x \cdot e^{x^2}}
$$
The second limit is again evaluated using L'Hospital:
$$
\lim \limits_{x \to \infty} \frac{G(x)}{2x \cdot e^{x^2}} = \frac{1}{2} \lim \limits_{x \to \infty} \frac{G'(x)}{e^{x^2} + 2 x^2 \cdot e^{x^2}} = \frac{1}{2} \lim \limits_{x \to \infty} \frac{e^{x^2}}{e^{x^2} + 2 x^2 \cdot e^{x^2}} = \frac{1}{2} \lim \limits_{x \to \infty} \frac{1}{1 + 2 x^2} = 0
$$
All in all, we arrive at
$$
\lim \limits_{x \to \infty} \int \limits_1^x x e^{t^2-x^2} dt = \lim \limits_{x \to \infty} \frac{x \cdot G(x)}{e^{x^2}} = \frac{1}{2}
$$
