Show that $\int_0^\infty e^{-(x-u/x)^2}(1-\frac{u}{x^2})dx$ converges uniformly on $u\in [\delta,L]$

Question

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?

Answer 1

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:

$$\int _{0}^{\infty }\text{exp}\left[- \left( x-{\frac {u}{x}} \right) ^{2}\right]\left( 1-{\frac {u}{{x}^{2}}} \right) {dx}=\lim_{\epsilon=0}\,\left[I_1(u,\epsilon)+I_2(u,\epsilon)\right]=0.$$

Answer 2

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.

Answer 3

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 $Q$.