If $f:\mathbb R^2\to \mathbb R$ continuous, s.t. $\int_{\mathbb R}f(t,x)dx$ exist for all $t$. Does $$\lim_{h\to 0}\int_0^{\infty } f(t+h,x)dx=\int_0^\infty f(t,x)dx \ \ ?$$

I asked here the question for $$\lim_{h\to 0}\int_a^b f(t+h,x)dx=\int_a^b f(t,x)dx,$$ where $a,b\in\mathbb R$ and as @Surb show, it's indeed true. But what happen when $a=-\infty $ and $b=\infty $ ? I think it's enough for $a=0$ and $b=\infty $, i.e. does

$$\lim_{h\to 0 }\int_0^\infty f(t+h,x)dx=\int_0^\infty f(t,x)dx \ \ ?$$ (we suppose of course that $\int_0^\infty f(t,x)dx$ exist for all $t$).

My idea was to show that $$F(v,t):=\int_0^v f(t,x)dx,$$ is continuous on $[0,\infty )\times \mathbb R$, and thus $$\begin{align}\lim_{h\to 0}\int_0^\infty f(t,x)dx=&\lim_{h\to 0}\lim_{v\to \infty }\int_0^v f(t+h,x)dx\\ &=\lim_{v\to \infty }\lim_{h\to 0}\int_0^v f(t+h,x)dx\\ &=\lim_{v\to \infty }\int_0^v f(t,x)dx\\ &=\int_0^\infty f(t,x)dx\end{align}$$

but unfortunately, I didn't success to show that it's continuous. May be it's not correct. And in this case, do you have a counter example ?

  • $\begingroup$ You say $f$ is defined on $\mathbb{R}$, but its argument has two variables. $\endgroup$ – Lorenzo Quarisa Jul 17 '18 at 8:57
  • $\begingroup$ What is $f(x,y)$ for a function from the real line into itself? Have you thought about existence of the integrals involved? Continuous functions are not necessarily integrable on an infinite interval. The question looks too vague in the present form. $\endgroup$ – Kavi Rama Murthy Jul 17 '18 at 8:59
  • $\begingroup$ @LorenzoQuarisa: I corrected it. $\endgroup$ – Peter Jul 17 '18 at 8:59
  • $\begingroup$ @KaviRamaMurthy: Yes, I add that $\int_{\mathbb R}f(x,t)dx$ exist for all $t$. $\endgroup$ – Peter Jul 17 '18 at 9:00
  • $\begingroup$ Your argument of $F=F(v,t)$ continuous doesn't allow you to conclude. For example $$g:(x,y)\longmapsto \begin{cases}\frac{x^3y^2}{x^2+y^2}&(x,y)\neq 0\\ 0&(x,y)=0\end{cases}$$ is continuous on $\mathbb R^2$ but $ \lim_{x\to \infty }\lim_{y\to 0}g(x,y)=0\neq \infty =\lim_{y\to 0}\lim_{x\to \infty }g(x,y)$. $\endgroup$ – Surb Jul 17 '18 at 9:04

Consider $f : \mathbb{R}^2 \to \mathbb{R}$ given by $f(t,x) = t^2e^{-t^2x}$.

We have $$\int_0^\infty f(0,x)\,dx = \int_0^\infty 0\,dx = 0$$


$$\lim_{h\to0} \int_0^\infty f(h,x)\,dx = \lim_{h\to0} \int_0^\infty h^2e^{-h^2x}\,dx = \lim_{h\to 0} \left[-e^{-h^2x}\right]_0^\infty = 1$$


This is not true: we will define a function $f(t, x)$ by describing it as a function of $x$ for a "fixed" $t$. For $t = 0$, we set $f(0, x) = 0$. For $t > 0$, we let the function $x \mapsto f(t, x)$ take the value 0 whenever $|x - 1/t| \geq 1$, and we set $f(t, 1/t) = 1$, and we complete the function $x \mapsto f(t, x)$ by making it piecewise linear: starting at $x = 1/t - 1$, the value of $f$ linearly increases to 1 as $x$ goes to $1/t$, and then it linearly descends to $0$ as $x$ goes to $1/t + 1$. Finally, to extend the function to $t < 0$, we set $f(t, x) = f(-t, x)$.

You can check that this function is continuous. Furthermore, trivially for all $t$ with $0 < |t| < 1$ we have $$ \int_0^\infty f(t, x)\,\mathrm dx = 1, $$ so that $$ \lim_{h \to 0}\int_0^\infty f(h, x)\,\mathrm dx = 1 \neq 0 = \int_0^\infty f(0, x) \,\mathrm dx. $$

  • $\begingroup$ Thanks for the counter example. May be you know a counter example with more elementary function ? (I would prefer to have a function a not construct a function if it's possible). $\endgroup$ – Peter Jul 17 '18 at 9:36
  • $\begingroup$ If you are willing to accept an integral that diverges to infinity instead of integrates to 1, then you can use $f(x, t) = xt$. Otherwise I can't think of a neat example immediately. $\endgroup$ – Mees de Vries Jul 17 '18 at 9:39

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