Some kind of relation between classical heat equation and Laplace .  If we have $k(x,t)= \frac {1}{(4t)^{\frac{n}{2}}} \exp\left(\frac{-|x|^2}{4t}\right)$ is the fundamental solution of heat equation. If we consider $n \ge 3 $, I would like to show that $\int_0^\infty k(x,t) dt$ is the fundamental solution of lLaplace equation. 
I would like some hints. 
I thought of integrating but don't know how to approach. 
Thank you 
ie , i need to arrive to a form like $\frac{1}{B} \frac{1}{|x|^{n-1}}$ $B $ is a constant depending on the measure of the space. 
 A: We use the substitution $s=\frac t{|x|^2}$ (then $dt=|x|^2ds$) to get 
\begin{align}
\int_0^{+\infty}k(x,t)dt&=\int_0^{+\infty}\frac 1{(4t)^{n/2}}\exp\left(-\frac{|x|^2}{4t}\right)dt\\
&=\frac 1{4^{n/2}}\int_0^{+\infty}\frac 1{(s|x|^2)^{n/2}}\exp\left(-\frac 1{4s}\right)|x|^2ds\\
&=\frac 1{4^{n/2}}|x|^{2-n}\int_0^{+\infty}\frac 1{s^{n/2}}\exp\left(-\frac 1{4s}\right)ds\\
&=\frac 1{4^{n/2}}|x|^{2-n}\int_0^{+\infty}y^{n/2-2}\exp(-y/4)dy\\
&=c_n|x|^{2-n}.
\end{align}
We have to show that $f\colon x\mapsto |x|^{2-n}=\left(\sum_{k=1}^nx_k^2\right)^{1-n/2}$ is harmonic. Let $j\in\{1,\dots,n\}$. We have 
$$\partial_jf(x)=\left(\sum_{k=1}^nx_k^2\right)^{-n/2}2x_j\left(1-\frac n2\right)$$
and 
$$\partial_{jj}f(x)=2\left(1-\frac n2\right)\left(\sum_{k=1}^nx_k^2\right)^{-n/2}-\left(1-\frac n2\right)nx_j\left(\sum_{k=1}^nx_k^2\right)^{-n/2-1}(2x_j).$$
Summing that, we get 
\begin{align}
\Delta f(x)&=\sum_{j=1}^n\partial_{jj}f(x)\\
&=\left(1-\frac n2\right)\left(\sum_{k=1}^nx_k^2\right)^{-n/2-1}\left(2n|x|^2-n\cdot 2\cdot |x|^2\right)\\
&=0.
\end{align}
A: Let $e(x)=\int_0^\infty k(x,t)\,\mathrm{d}t$, and let us compute the action of the distribution $\Delta e$ on the test function $\phi$ (which is a compactly supported smooth function in $\mathbb{R}^n$) as
$$
\begin{split}
\langle\Delta e,\phi\rangle 
&= \int_{\mathbb{R}^n} e(x) \Delta\phi(x)\,\mathrm{d}x 
= \int_{\mathbb{R}^n} \int_0^\infty k(x,t) \Delta\phi(x)\,\mathrm{d}t\,\mathrm{d}x 
= \int_0^\infty \int_{\mathbb{R}^n} k(x,t) \Delta\phi(x)\,\mathrm{d}x\,\mathrm{d}t \\
&= \int_0^\infty \int_{\mathbb{R}^n} \Delta k(x,t) \phi(x)\,\mathrm{d}x\,\mathrm{d}t
= \int_0^\infty \int_{\mathbb{R}^n} \partial_t k(x,t) \phi(x)\,\mathrm{d}x\,\mathrm{d}t \\
&= \int_0^\infty \int_{\mathbb{R}^n} \partial_t k(x,t) \phi(x)\,\mathrm{d}x\,\mathrm{d}t 
= -\lim_{t\to0}\int_{\mathbb{R}^n} k(x,t) \phi(x)\,\mathrm{d}x 
= - \phi(0),
\end{split}
$$
where we have used the fact that $\Delta k =\partial_tk$ and that $\int_{\mathbb{R}^n} k(x,t) \phi(x)\,\mathrm{d}x\to\phi(0)$ as $t\to0$.
Thus we have $\Delta e = - \delta$. Some would say that $e$ is not exactly a fundamental solution, but that $-e$ is one.
