Are there known expressions for total variation distance between $N(0,\sigma_1^2)$ and $N(0,\sigma^2)$ Are known expressions for total variation distance between $N(0,\sigma^2)$ and $N(0,\sigma^2+\epsilon)$ for small $\epsilon$? The only thing I seem to find is things are expression about the mean but not if we change variance slightly.
 A: There's also a softer argument based on properties of the heat kernel, which applies in higher dimensions as well, in Lemma 4.9 of this paper of Klartag.  It shows that in $n$ dimensions, the total variation distance between centered Gaussian distributions with covariances $\alpha I_n$ and $\beta I_n$ is at most
$$
C \sqrt{n} \left|\frac{\beta}{\alpha} - 1\right|,
$$
where $C > 0$ is an absolute constant (which can be made explicit if you want).
A: For two measures with densities f,g, the total variation distance is
$$
\int_{f>g}(f(x)-g(x))\,dx
$$
For two Gaussian measures with the same mean and different variances it is easy to identify the set $\{f>g\}$ and to obtain the formula for the distance in terms of the partition function of the standard Gaussian measure
A: As marcoromito wrote, this is an elementary calculation.  However, I thought I would record a nice approximation that I stumbled across.  Whether it is new, I have no idea.
ADDED: The following sentence has changed.
By scaling, we can restrict ourselves to $\mathrm{tvd}(\epsilon)$ which is the total variation distance between $N(0,1)$ and $N(0,(1+\epsilon)^2)$ for $\epsilon\ge 0$.  The points
where the two densities are equal are $\pm x_0$ where
$$x_0 = \frac{(1+\epsilon)\sqrt{2\ln(1+\epsilon)}}{\sqrt{\epsilon(2+\epsilon)}}.$$
Integrating the difference between the densities over $[-x_0,x_0]$ we find
$$\mathrm{tvd}(\epsilon) = 
  \mathrm{erf}\left( \frac{(1+\epsilon)\sqrt{\ln(1+\epsilon)}}{\sqrt{\epsilon(2+\epsilon)}}\right)
- \mathrm{erf}\left( \frac{\sqrt{\ln(1+\epsilon)}}{\sqrt{\epsilon(2+\epsilon)}}\right).
$$
The interesting thing is that for small $\epsilon\ge 0$
$$ \mathrm{tvd}(\epsilon) = \mathrm{tvd'}(\epsilon)
 + O(\epsilon^5)$$
where
$$ \mathrm{tvd'}(\epsilon) = \frac{2^{3/2}\,\epsilon }{\pi^{1/2}e^{1/2}(2+\epsilon)}.$$
This rational approximation is remarkably precise for small $\epsilon$.
Experimentally (based on plotting the ratio in Maple and seeing a smooth curve),
$$1 \le \frac{\mathrm{tvd}(\epsilon)}{\mathrm{tvd'}(\epsilon)}
 \le \sqrt{e\pi/8} \approx 1.03318$$ for all $\epsilon\gt 0$.
