Perimeter of circumference using Laplace transform I wish to compute the perimeter of a circumference of unit radius by using the Laplace transform (just for fun). However, I am getting the wrong result by a factor of 2. Can someone help me find it?
I write $$P=\int dx dy \delta(x^2+y^2-1),$$ where the integral is over the whole plane. Then I introduce the auxiliary function $$P(t)=\int dx dy \delta(x^2+y^2-t).$$
The Laplace transform $f(s)=\int_0^\infty e^{-st}P(t)dt$ gives 
$$f(s)= \int dx dy e^{-s(x^2+y^2)}=2\pi\int_0^\infty rdre^{-sr^2}=\frac{\pi}{s}.$$
Inverse Laplace transform then gives just $P(t)=\pi$. 
Where is the mistake?
 A: You don't need the Laplace transform to see the problem. Notice that 
$$\int dxdy\delta(x^2+y^2-1)=2\pi\int_0^\infty rdr\delta(r^2-1)=2\pi\frac{1}{2}=\pi.$$
This is wrong because you should have $\delta(r-1)$ in there and not $\delta(r^2-1)$ (user md2perp2 has it right on his comment).
So you should start with $$P=\int dxdy\delta(\sqrt{x^2+y^2}-1)$$ and then it will work.
A: Here it is a more rigorous way. Let $D_\rho$ the disk centered at the origin with radius $\rho$ and $L$ the length of the unit circle. Since $\partial D_\rho$ is smooth we have
$$ L = \lim_{\varepsilon\to 0}\frac{\mu(D_{\rho+\varepsilon})-\mu(D_\rho)}{\varepsilon} \tag{1}$$
but on the other hand $\mu(D_\rho)=C\rho^2$, hence $L=2\mu(D_1)$. By Fubini's theorem
$$ \iint_{\mathbb{R}^2}e^{-(x^2+y^2)}\,dx\,dy = \left(\int_{-\infty}^{+\infty}e^{-z^2}\,dz\right)^2 \tag{2} $$
and by parity and the Laplace transform
$$ \int_{-\infty}^{+\infty}e^{-z^2}\,dz = 2\int_{0}^{+\infty}e^{-z^2}\,dz = \int_{0}^{+\infty}\frac{dx}{e^x\sqrt{x}}=\left(\mathcal{L}\frac{1}{\sqrt{x}}\right)(1)=\sqrt{\pi}.\tag{3} $$
On the other hand, by Cavalieri's principle the LHS of $(2)$ equals
$$ \int_{0}^{+\infty} L\rho e^{-\rho^2}\,d\rho = \frac{L}{2}\tag{4}$$
hence by $(2),(3)$ and $(4)$ we have $L=2\pi$ and $\mu(D_1)=\pi$.
A: I believe the problem lies in the use of the Delta Dirac distribution, a defining property of which states $$ \int_{-\infty}^{\infty} f(x) \delta(x) \mathrm{d}x = f(0)$$
In your case, after switching the limits of integration (a step itself that might require some justification), you get an integral over $t$ between $0$ and $\infty$  and not over the real line as the Delta Dirac distribution would require.
I think I saw a post here on MSE where the “audacious” pushed it forward, up to claiming $$ \int_{0 }^{\infty} f(x) \delta(x) \mathrm{d}x = \frac{f(0)}{2}$$  explained to the laymen as myself by considering nascent delta functions symmetric with respect to the origin and observing their limit. I am sure one could find calcualtions where such approach operationally works, the kind of pindaric flights physicists often delight audiences with.
In this case, whose rigorous validity I am totally unqualified to comment upon, one might be tempted to think you are really computing only half of the circumference, which fits your result.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
P & \equiv
\iint_{\large\mathbb{R}^{2}}\delta\pars{x^{2} + y^{2} - 1}\dd x\,\dd y
\\[5mm] & =
\int_{-\infty}^{\infty}\bracks{\verts{y} < 1}\int_{-\infty}^{\infty}
\bracks{{\delta\pars{x + \root{1 - y^{2}}} \over \verts{2x}} +
{\delta\pars{x - \root{1 - y^{2}}} \over \verts{2x}}}\dd x\,\dd y
\\[5mm] & =
\int_{-1}^{1}\bracks{%
{1 \over 2\root{1 - y^{2}}} + {1 \over 2\root{1 - y^{2}}}}\,\dd y
\,\,\,\stackrel{y\ =\ \sin\pars{\theta}}{=}\,\,\,\int_{-\pi/2}^{\pi/2}\dd\theta = \bbx{\pi}
\end{align}
