Show that $\int_0^\infty e^{-(x-u/x)^2}(1-\frac{u}{x^2})dx$ converges uniformly on $u\in [\delta,L]$ I would like to show that $$\int_0^\infty e^{-(x-u/x)^2}\left(1-\frac{u}{x^2}\right)dx$$ converges uniformly on, say, $u\in [\delta,L]$, for arbitrarily small $\delta>0$ and arbitrarily large $L>0$. The important thing is that we can fit any $u>0$ into some interval on which the integral converges uniformly.
If we can major the exponential by some appropriate exponential in terms of $x$, then the rest is taken care of. But I am having a hard time doing so. For instance, I may try $e^{-(x-u/x)^2}\leq e^{-x^2}$, but it is not clear that this is true for instance when both $u$ and $x$ are close to zero.
Any ideas?
 A: Let $0 < \delta <L$. Given $\varepsilon \in \Bbb{R}_{>0}$, we want
\begin{split}
\left \vert\int_0^{\frac{1}{n}} e^{-(x-u/x)^2}\left(1-\frac{u}{x^2}\right)dx+\int_n^\infty e^{-(x-u/x)^2}\left(1-\frac{u}{x^2}\right)dx \right\vert < \varepsilon
\end{split}
for all big enough $n$ (i.e. all $n \geq N$ for some $N \in \Bbb{N}$) and all $u \in [\delta,L]$.
First, note that for $1/n^2< \delta$, we have
\begin{split}
\left \vert\int_0^{\frac{1}{n}} e^{-(x-u/x)^2}\left(1-\frac{u}{x^2}\right)dx \right \vert&=\left \vert\int_0^{\frac{1}{n}} e^{-x^2(1-u/x^2)^2}\left(1-\frac{u}{x^2}\right)dx \right \vert\\
&=\left \vert\int_0^{\frac{1}{n}} \frac{1}{\frac{1}{1-u/x^2}+x\sum_{j=1}^{\infty}\frac{(x-u/x)^{2j-1}}{j!}}dx \right \vert\\
&=\int_0^{\frac{1}{n}} \frac{1}{\underbrace{\frac{1}{u/x^2-1}}_{\geq 0}+\underbrace{x\sum_{j=1}^{\infty}\frac{(u/x-x)^{2j-1}}{j!}}_{\geq \delta-x^2}}dx\\
&\leq \int_0^{\frac{1}{n}} \frac{1}{\delta-x^2}dx\\
&< \varepsilon/2
\end{split}
with the last inequality holding for all big enough $n$.
Similarly, we have for $n^2>L$
\begin{split}
\left \vert\int_n^\infty e^{-(x-u/x)^2}\left(1-\frac{u}{x^2}\right)dx \right \vert&=\int_n^\infty \frac{1}{\underbrace{\frac{1}{1-u/x^2}}_{\geq 0}+\underbrace{x\sum_{j=1}^{\infty}\frac{(x-u/x)^{2j-1}}{j!}}_{\geq x^2-L}}dx\\
&\leq \int_n^{\infty} \frac{1}{x^2-L}dx\\
&< \varepsilon/2
\end{split}
for all big enough $n$, and the result follows.
A: Weierstrass M-test
Let
$f(x,u) = e^{-(x-u/x)^2}(1-u/x^2)$.
It can be shown that $|f|$ is bounded by $1/(\sqrt{2 e}x)$.
For $u>0$, $\lim_{x\to 0^+}f(x,u) = 0$.
This, along with the above fact, implies that for any $u\in[\delta,L]$ there is an $x_0>0$ such that for $x<x_0$,
$|f|<1/(\sqrt{2e}x_0)$.
(For $\delta\ll 1$, $x_0\simeq \delta$.)
Now we show that
$|f(x,u)| < e^{-(x-\sqrt{L})^2}$ for $0<u\le L$ and $x>\sqrt{L}$,
that is for $u = (1-\alpha)L$ and $x = (1+\beta)\sqrt{L}$,
where $\alpha\in[0,1)$ and $\beta\in(0,\infty)$.
First notice that
$$|f(x,u)|
= e^{-(x-u/x)^2}|1-u/x^2|
< e^{-(x-u/x)^2}$$
since $u\le L < x^2$.
But
$$(x-u/x)^2 - (x-\sqrt{L})^2
= \frac{(\alpha+\beta)(2\beta^2+3\beta+\alpha)L}{(1+\beta)^2},$$
which is explicitly positive definite. 
This proves the claim. 
Therefore, $|f|$ is bounded for all $u\in[\delta,L]$
by the following integrable function,
$$g(x) =
\left\{\begin{array}{ll}
1/(\sqrt{2e}x_0), & x < x_0 \\
1/(\sqrt{2e}x), & x_0 \le x \le \sqrt{L} \\
e^{-(x-\sqrt{L})^2}, & x > \sqrt{L}.
\end{array}\right.$$
From the definition
As mentioned by @GrahamHesketh, the integral vanishes for any $u>0$.
Let $x=\frac{\sqrt{u}}{2}(t+\sqrt{t^2+4})$.
Then
$$\int_0^\infty e^{-(x-u/x)^2}(1-u/x^2)dx
= \sqrt{u}\int_{-\infty}^\infty \frac{t e^{-u t^2}}{\sqrt{t^2+4}}dt = 0.$$
Thus, we wish to show that for any $\epsilon>0$ we can find a $Q$ independent of $u$ such that when $R>Q$,
$$\left|\int_0^R f(x,u)dx\right| < \epsilon, \quad u\in[\delta,L].$$
But $\int_0^R = \int_0^\infty - \int_R^\infty = -\int_R^\infty$, and
$$\begin{eqnarray*}
\int_R^\infty f(x,u)dx
&<& \int_R^\infty e^{-(x-\sqrt{L})^2}dx,
\qquad (\textrm{assume }Q>\sqrt{L}) \\
&=& \frac{\sqrt{\pi}}{2}\mathrm{erfc}(R-\sqrt{L}),
\end{eqnarray*}$$
where $\mathrm{erfc}$ is the complementary error function.
Clearly for any $L$ we can make
$\frac{\sqrt{\pi}}{2}\mathrm{erfc}(R-\sqrt{L})$
as small as we wish by choosing a large enough $Q$.
A: Formally let us define the integral as:
$$\int _{0}^{\infty}\text{exp}\left[- \left( x-{\frac {u}{x}} \right) ^{2}\right]\left( 1-{\frac {u}{{x}^{2}}} \right) {dx}=\lim_{\epsilon=0}\int _{\epsilon}^{1/\epsilon }\text{exp}\left[- \left( x-{\frac {u}{x}} \right) ^{2}\right]\left( 1-{\frac {u}{{x}^{2}}} \right) {dx}.
$$
For $u>0$ rescale such that $x\rightarrow y\sqrt{u}$:
$$\int _{\epsilon}^{1/\epsilon }\text{exp}\left[- \left( x-{\frac {u}{x}} \right) ^{2}\right]\left( 1-{\frac {u}{{x}^{2}}} \right) {dx}=u^{\frac{1}{2}}\int _{\epsilon/\sqrt{u}}^{1/(\epsilon\sqrt{u}) }\text{exp}\left[- u\left( y-{\frac {1}{y}} \right) ^{2}\right]\left( 1-\dfrac{1}{y^2}\right){dy}
$$
then split the integral into two parts:
$$\sqrt {u}\int _{\epsilon/\sqrt{u}}^{\sqrt{u}/\epsilon }\text{exp}\left[- u\left( y-{\frac {1}{y}} \right) ^{2}\right]\left( 1-\dfrac{1}{y^2}\right) {dy}=I_1(u,\epsilon)+I_2(u,\epsilon),$$
$$I_1(u,\epsilon)=\sqrt {u}\int _{\epsilon/\sqrt{u}}^{1 }\text{exp}\left[- u\left( y-{\frac {1}{y}} \right) ^{2}\right]\left( 1-\dfrac{1}{y^2}\right) {dy},$$ $$I_2(u,\epsilon)=\sqrt {u}\int _{1}^{1/(\epsilon\sqrt{u}) }\text{exp}\left[- u\left( y-{\frac {1}{y}} \right) ^{2}\right]\left( 1-\dfrac{1}{y^2}\right) {dy}.
$$
Substituting $y=1/t$ in $I_1(u,\epsilon)$ leaves the integrand invariant up to a sign change and shows that:
$$\int _{\epsilon/\sqrt{u}}^{1 }\text{exp}\left[- u\left( y-{\frac {1}{y}} \right) ^{2}\right]\left( 1-\dfrac{1}{y^2}\right) {dy}=-\int _{1}^{\sqrt{u}/\epsilon }\text{exp}\left[- u\left( t-{\frac {1}{t}} \right) ^{2}\right]\left( 1-\dfrac{1}{t^2}\right) {dt},$$
from which it follows that:
$$\lim_{\epsilon=0}\,\left[I_1(u,\epsilon)+I_2(u,\epsilon)\right]=\lim_{\epsilon=0}\sqrt{u}\int _{\sqrt{u}/\epsilon}^{1/(\epsilon\sqrt{u}) }\text{exp}\left[- u\left( t-{\frac {1}{t}} \right) ^{2}\right]\left( 1-\dfrac{1}{t^2}\right) {dt}$$
where the integrand is monotonic and decreasing for $t>1$ and thus it is less than the smallest limit, so for $u\le1$:
$$ \left|\int _{\sqrt{u}/\epsilon}^{1/(\epsilon\sqrt{u}) }\text{exp}\left[- u\left( t-{\frac {1}{t}} \right) ^{2}\right]\left( 1-\dfrac{1}{t^2}\right) {dt}\right| <\text{exp}\left[-u \left( {\frac {\sqrt {u}}{
\epsilon}}-{\frac {\epsilon}{\sqrt {u}}} \right) ^{2}\right] \left( 1-{
\frac {{\epsilon}^{2}}{u}} \right)  \left( {\frac {1}{\sqrt {u}\epsilon}}-{\frac {\sqrt {u}}{\epsilon
}} \right)
 $$
and for $u\ge 1$:
$$ \left| \int _{1/(\epsilon\sqrt {u})}^{\sqrt{u}/\epsilon}\text{exp}\left[- u\left( t-{\frac {1}{t}} \right) ^{2}\right]\left( 1-\dfrac{1}{t^2}\right) {dt}\right| <\text{exp}\left[-u \left( {\dfrac {1}{\sqrt {u}
\epsilon}}-\sqrt {u}\epsilon \right) ^{2}\right] \left( 1-u{\epsilon}^{2}
 \right)  \left( {\dfrac {\sqrt {u}}{\epsilon}}-{\dfrac {1}{\sqrt {u}
\epsilon}} \right)$$ 
both of which are Gaussian dominated and vanish as $\epsilon\rightarrow 0$ independently of $u$. It follows that: $$\forall \,u,\,\sigma>0,\,\exists\, \epsilon\,\,\, \text{such that:}\,\, I_1(u,\epsilon)+I_2(u,\epsilon)<\sigma,$$ (uniform convergence) and furthermore:
$$\int _{0}^{\infty }\text{exp}\left[- \left( x-{\frac {u}{x}} \right) ^{2}\right]\left( 1-{\frac {u}{{x}^{2}}} \right) {dx}=\lim_{\epsilon=0}\,\left[I_1(u,\epsilon)+I_2(u,\epsilon)\right]=0.$$

