How to find $\lim\limits_{t \to \infty} \int\limits_0^\infty u(x,t) dx$? I have $u(x,t)$ - solution of Cauchy problem $$u_t=u_{xx},~~~u(x,0)=e^{-x^2},$$ where $t>0, x\in\mathbb{R}.$
Is there a way to find such a limit? I have seen several theorems which help to find value of $\lim\limits_{t \to \infty} u(x,t),$ but this integral realy spoils everything.
Thank you.
 A: The $L_p$ solution to the problem is though convolution with Gauss-Wierstrass kernel,
$$ 
u(x,t)= W_t*f(x)=\frac{1}{\sqrt{4\pi t}}\int f(y)e^{\frac{-|x-y|^2}{4t}}\,dy
$$
where $f(x)=e^{-x^2}$. The kernel $W_t$ is a good approximation to the identity and so, $W_t*f\rightarrow f$ in $L_p$ ($1\leq p<\infty$ ) and point wise to $f$ (it should be at any Lebesgue point of $f$ fut $f(x)=e^{-x^2}$ is continuous so every point in $\mathbb{R}$ is a Lebesgue point of $f$. In fact, the problem is simple here since $f(x)$ is uniformly continuous.
Here are some generalities of approximation to identity. You have to show that $W_t(x)=t^{-1}W_1(x/t)$, where $W_1(y)=\frac{1}{\sqrt{4\pi}}e^{-y^2}$ satisfies the conditions (1)--(3) described below:

Consider a collection $\{K_\varepsilon:\varepsilon>0\}\subset\mathcal{L}_1(\mathbb{R}^n,\lambda_n)$ that satisfy the following properties:


*

*$\int_{\mathbb{R}^n} K_\varepsilon(x)\,dx=a$ for all
$\varepsilon>0$.

*$\sup_{\varepsilon>0}\|K_\varepsilon\|_1<\infty$.

*$\int_{|x|>\delta}|K_\varepsilon(x)|\,dx\rightarrow0$ as
$\varepsilon\rightarrow0$.


Theorem.
  Suppose $\{K_\varepsilon:\varepsilon>0\}\subset\mathcal{L}_1(\mathbb{R}^n,\lambda_n)$ satisfy (1)--(3) above. Then, for any  $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$, $1\leq p<\infty$,
\begin{aligned}
\lim_{\varepsilon\rightarrow0}\|f* K_{\varepsilon} - a\,f\|_p = 0.
\end{aligned}
If $f\in\mathcal{L}_\infty(\mathbb{R}^n,\lambda_n)$ is continuous at a point $x$, then
$\lim_{\varepsilon\rightarrow0}f*K_\varepsilon(x)=f(x)$. If  $f$ is bounded and uniformly continuous, then $f*K_\varepsilon$ converges to $f$ uniformly as $\varepsilon\rightarrow0$.
Proof: 
Let $M=\sup_{\varepsilon>0}\|K_\varepsilon\|_1$. If $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$, $1\leq p<\infty$, and application of the generalized Minkowski's inequality gives
\begin{aligned}
\|f* K_\varepsilon -af\|_p &\leq  \Big(\int\Big(\int_{\mathbb{R}^n}|f(x-y)-f(x)|
|K_\varepsilon(y)|\,dy\Big)^p\,dx\Big)^{1/p}\\
& \leq \int_{\mathbb{R}^d}\Big(\int|f(x-y)-f(x)|^p\,dx\Big)^{1/p}|K_\varepsilon(y)|\,dy\\
&=\int\|\tau_yf-f\|_p|K_\varepsilon(y)|\,dy.
\end{aligned}
where $\tau_y$ is the translation operator. Recall that $\lim_{y\rightarrow0}\|\tau_yf-f\|_p=0$ for all $f\in\mathcal{L}_p$. Assumption (2) implies that for any $\eta>0$,  there exists $\delta>0$ such that $M\|\tau_yf-f\|_p<\eta/2$ whenever $|y|\leq\delta$. By assumption (3), for some $\varepsilon'>0$ we have
$2\|f\|_p\int_{|y|>\delta}|K_\varepsilon(x)|\,dx<\eta/2$ whenever $\varepsilon<\varepsilon'$. Combining these facts,  we obtain
\begin{aligned}
\|f*K_\varepsilon-af\|_p\leq&
\int_{|y|\leq\delta}\|\tau_yf-f\|_p|K_\varepsilon(y)|\,dy\\
&\quad + \int_{|y|>\delta}\|\tau_yf-f\|_p|K_\varepsilon(y)|\,dy
\leq \frac{\eta}{2} + \frac{\eta}{2}
\end{aligned}
whenever $0<\varepsilon<\varepsilon'$.
The second statement follows similarly. Let $\eta>0$ be fixed. If $f$ is continuous at $x$, then for some $\delta>0$, $|x-u|\leq\delta$ implies that 
$|f(x)-f(u)|<\frac{\eta}{2 M}.$
 For such $\delta>0$, there is $\varepsilon'>0$ such that
$2\|f\|_{\infty}\int_{|x|>\delta}|K_\varepsilon(x)|\,dx<\frac{\eta}{2}$ whenever $0<\varepsilon<\varepsilon'$. Putting these statements together gives
\begin{aligned}
|f*K_\varepsilon(x)&-af(x)|\leq \int |f(x-y)-f(x)||K_\varepsilon|(y)\,dy\\
&\leq 
\int_{|y|\leq\delta}+\int_{|y|>\delta}|f(x-y)-f(x)||K_\varepsilon|(y)\,dy
\leq \frac{\eta}{2}+\frac{\eta}{2}
\end{aligned}
If $f$ is bounded and uniformly continuous, then $\delta>0$ can be chosen so that 
$$\sup_{|v-u|<\delta}|f(u)-f(u)|<\frac{\eta}{2M}.$$
Uniform convergence follows immediately.
A: I believe you have that:
$$u(x,t)=e^{-x^2}f(t)$$
and so you want:
$$\lim_{t\to\infty}f(t)\int_0^\infty e^{-x^2}dx$$
assuming that this is interchangeable goes to:
$$\frac{\sqrt{\pi}}{2}\lim_{t\to\infty}f(t)$$
