finding unknown variable in Gaussian Integral Given values of d, p and $\sigma$, is it possible to calculate the value of $\mu$?
$$1-\frac{1}{2\pi\sigma^2}\int_{-\infty}^{\infty}\int_{y-d}^{y+d}\exp\big(-{x^2}/{2\sigma^2}\big) \exp\big(-{(y-\mu)^2}/{2\sigma^2}\big) \,\mathrm{d}x\,\mathrm{d}y < p$$
 A: Your double integral can be evaluated in closed form. This is done by evaluating $x$ integral first, differentiating with respect to $d$, and remembering that for $d=0$ the integral vanishes. Then carrying out $y$ integration, and then integration with respect to $d$ from zero to $d$. Using Mathematica:
In[19]:= 1 - 
 Integrate[
  Integrate[
   D[1/(2 Pi si^2)
       Integrate[
       Exp[-x^2/(2 si^2)] Exp[-(y - mu)^2/(2 si^2)], {x, y - dd, 
        y + dd}], dd] // FullSimplify, {y, -Infinity, Infinity}, 
   Assumptions -> si > 0], {dd, 0, d}]

Out[19]= 1 + 1/2 (-Erf[(d - mu)/(2 si)] - Erf[(d + mu)/(2 si)])

Hence your problem becomes $$1-\frac{1}{2} \left( \text{erf}\left( \frac{\mu+d}{2\sigma} \right) +  \text{erf}\left( \frac{d-\mu}{2\sigma} \right) \right) < p$$.
From this inequality one may deduce the implies inequality for $\mu$. Here is an example:

A: Let $I$, as in my first answer, denote the iterated integral (including the factor $\frac{1}{{2\pi \sigma ^2 }}$). Using probabilistic arguments, I obtained
$$
I = \Phi \bigg(\frac{{\mu  + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu  - d}}{{\sqrt {2\sigma^ 2} }}\bigg),
$$
where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution.
On the other hand, using Mathematica, Sasha obtained 
$$
I = \frac{1}{2}\bigg({\rm erf}\bigg(\frac{{\mu  + d}}{{2\sigma }}\bigg) + {\rm erf}\bigg(\frac{{d - \mu }}{{2\sigma }}\bigg)\bigg).
$$ 
Here ${\rm erf}$ is the error function, defined by
$$
{\rm erf}(x) = \frac{2}{{\sqrt \pi  }}\int_0^x {e^{ - t^2 } \,dt} , \;\; x \in \mathbb{R}.
$$
(Note that ${\rm erf}(-x)=-{\rm erf}(x)$.)
So, let's show that indeed
$$
\Phi \bigg(\frac{{\mu  + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu  - d}}{{\sqrt {2\sigma^ 2} }}\bigg) = \frac{1}{2}\bigg({\rm erf}\bigg(\frac{{\mu  + d}}{{2\sigma }}\bigg) + {\rm erf}\bigg(\frac{{d - \mu }}{{2\sigma }}\bigg)\bigg).
$$
From the standard relation
$$
\Phi (x) = \frac{1}{2}\bigg[1 + {\rm erf}\bigg(\frac{x}{{\sqrt 2 }}\bigg)\bigg],
$$
we get
$$
\Phi \bigg(\frac{{\mu  + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu  - d}}{{\sqrt {2\sigma^ 2} }}\bigg) = 
\frac{1}{2}\bigg({\rm erf}\bigg(\frac{{\mu  + d}}{{2\sigma }}\bigg) - {\rm erf}\bigg(\frac{{\mu  - d}}{{2\sigma }}\bigg)\bigg),
$$
and hence the desired equality follows from 
$$
{\rm erf}\bigg(\frac{{d - \mu }}{{2\sigma }}\bigg) =  - {\rm erf}\bigg(\frac{{\mu  - d}}{{2\sigma }}\bigg).
$$
Now, as in my first answer, define a function $f$ by
$$
f(\mu):=\Phi \bigg(\frac{{\mu  + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu  - d}}{{\sqrt {2\sigma^ 2} }}\bigg) = 
\frac{1}{2}\bigg({\rm erf}\bigg(\frac{{\mu  + d}}{{2\sigma }}\bigg) - {\rm erf}\bigg(\frac{{\mu  - d}}{{2\sigma }}\bigg)\bigg).
$$
Recall that $f$ is decreasing in $\mu \in [0,\infty)$, with $f(\mu) \to 0$ as $\mu \to \infty$.
So if $f(0) > 1-p$, there exists a solution $\mu > 0$ to $f(\mu)=1-p$. You can find an extremely accurate approximation to $\mu$ using, for example, Wolfram Alpha (based on the representation using the error function). 
A: It can be easily shown (using the law of total probability)* that
$$
\frac{1}{{2\pi \sigma ^2 }}\int_{ - \infty }^\infty  {\int_{ y-d }^{y+d}  {\exp \bigg( - \frac{{x^2 }}{{2\sigma ^2 }}\bigg)\exp \bigg( - \frac{{(y - \mu )^2 }}{{2\sigma ^2 }}\bigg) {\rm d}x} \,{\rm d}y} = \Phi \bigg(\frac{{\mu  + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu  - d}}{{\sqrt {2\sigma^ 2} }}\bigg),
$$
where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution.
Noting that the right-hand side is maximized when $\mu = 0$ (indeed, consider the integral of the ${\rm N}(0,1)$ pdf over the fixed length interval $[\frac{{\mu  - d}}{{\sqrt {2\sigma ^2 } }},\frac{{\mu  + d}}{{\sqrt {2\sigma ^2 } }}]$), it follows that a necessary condition for your inequality to hold is
$$
\Phi \bigg(\frac{{d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{-d}}{{\sqrt {2\sigma^ 2} }}\bigg) > 1 - p.
$$
On the other hand, if this condition is satisfied, then your inequality holds with $\mu=0$.
To summarize: The inequality holds for some $\mu \in \mathbb{R}$ if and only if it holds for $\mu=0$; the inequality for $\mu = 0$ is equivalent to
$$
\Phi \bigg(\frac{{d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{-d}}{{\sqrt {2\sigma^ 2} }}\bigg) > 1 - p.
$$
EDIT (in view of your comment below Sasha's answer): Assume that the necessary condition above is satisfied. The function $f$ defined by
$$
f(\mu ) = \Phi \bigg(\frac{{\mu  + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu  - d}}{{\sqrt {2\sigma^ 2} }}\bigg)
$$
is decreasing in $\mu \in [0,\infty)$, with $f(\mu) \to 0$ as $\mu \to \infty$. By our assumption, $f(0) > 1-p$. So if you are interested in a $\mu > 0$ such that $f(\mu) \approx 1-p$, you need to find $\mu_1,\mu_2 > 0$ such that $f(\mu_1) > 1- p$ and $f(\mu_2) < 1-p$, and $f(\mu_1) - f(\mu_2) \approx 0$. Then, for any $\mu \in (\mu_1,\mu_2)$, $f(\mu) \approx 1-p$.
* EDIT: Derivation of the first equation above. Denote the left-hand side of that equation by $I$.
First write $I$ as
$$
I = \int_{ - \infty }^\infty  {\bigg[\int_{y - d}^{y + d} {\frac{1}{{\sqrt {2\pi \sigma ^2 } }}\exp \bigg( - \frac{{x^2 }}{{2\sigma ^2 }}\bigg){\rm d}x} \bigg]\frac{1}{{\sqrt {2\pi \sigma ^2 } }}\exp \bigg( - \frac{{(y - \mu )^2 }}{{2\sigma ^2 }}\bigg){\rm d}y} .
$$
Then
$$
I = \int_{ - \infty }^\infty  {{\rm P}( - d \le X - y \le d)\frac{1}{{\sqrt {2\pi \sigma ^2 } }}\exp \bigg( - \frac{{(y - \mu )^2 }}{{2\sigma ^2 }}\bigg){\rm d}y} ,
$$
where $X$ is a ${\rm N}(0,\sigma^2)$ random variable. If $Y$ is a ${\rm N}(\mu,\sigma^2)$ random variable independent of $X$, then, by the law of total probability,
$$
{\rm P}( - d \le X - Y \le d) = \int_{ - \infty }^\infty  {{\rm P}( - d \le X - Y \le d|Y = y)f_Y (y)\,{\rm d}y} = I,
$$
where $f_Y$ is the pdf of $Y$, given by 
$$
f_Y (y) = \frac{1}{{\sqrt {2\pi \sigma ^2 } }}\exp \bigg( - \frac{{(y - \mu )^2 }}{{2\sigma ^2 }}\bigg),
$$
and where for the last equality ($\int_{ - \infty }^\infty   \cdot =I$) we also used the independence of $X$ and $Y$. Now, $X-Y \sim {\rm N}(-\mu,2\sigma^2)$; hence
$$
\frac{{(X - Y) - ( - \mu )}}{{\sqrt {2\sigma ^2 } }} \sim {\rm N}(0,1),
$$
and, in turn,
$$
I = {\rm P}\bigg(\frac{{ - d - ( - \mu )}}{{\sqrt {2\sigma ^2 } }} \le Z \le \frac{{d - ( - \mu )}}{{\sqrt {2\sigma ^2 } }}\bigg) = {\rm P}\bigg(\frac{{\mu  - d}}{{\sqrt {2\sigma ^2 } }} \le Z \le \frac{{\mu  + d}}{{\sqrt {2\sigma ^2 } }}\bigg),
$$
where $Z \sim {\rm N}(0,1)$. Thus, finally, 
$$
I = \Phi \bigg(\frac{{\mu  + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu  - d}}{{\sqrt {2\sigma^ 2} }}\bigg).
$$
