I am interested in the following integral $$\int_{-a}^adx\int_{-a}^a\mathop{\mathrm{d}y}xy\frac{1}{\sqrt{b}}\exp\left[-\frac{(x-y)^2}{2b}\right]\sqrt{\frac{2}{c}}\exp\left[-\frac{(x+y)^2}{4c}\right].$$ Does any one know how to evaluate the above integral? Mathematica was not able to give result for this.

  • $\begingroup$ Well I think it's $0$ for $c=b/2$, because it becomes an odd function in both $x$ and $y$. $\endgroup$ – snulty Jun 23 '16 at 21:42
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    $\begingroup$ Try $u=x+y\, ,v=x-y$ the integrals decouple into pieces which can be expresed in terns of error functions $\endgroup$ – tired Jun 23 '16 at 21:54
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    $\begingroup$ A bit a context? What is the reason for computing the probability that a normal random variable takes values over a rhombus centered at the origin? $\endgroup$ – Jack D'Aurizio Jun 23 '16 at 22:25
  • $\begingroup$ It is to compute some correlations. $\endgroup$ – titanium Jun 24 '16 at 11:22

Let's make this integral a little nicer. First, we deal with the parameters:


$$x=au, \qquad y=av, \qquad b=a^2 B, \qquad c=\frac{a^2 C}{2}$$

$$=\frac{2a^2}{\sqrt{BC}} \int_{-1}^1 \int_{-1}^1 u~v~\exp \left[-\frac{(u-v)^2}{2B} \right]~\exp \left[-\frac{(u+v)^2}{2C} \right]\mathrm{d}u~\mathrm{d}v$$

Let's forget about the coefficients and name the integral:

$$I_0(B,C)=\int_{-1}^1 \int_{-1}^1 u~v~\exp \left[-\frac{(u-v)^2}{2B} \right]~\exp \left[-\frac{(u+v)^2}{2C} \right]\mathrm{d}u~\mathrm{d}v$$

Separating the region into four obvious parts and introducing another integral:

$$I(B,C)=\int_0^1 \int_0^1 u~v~\exp \left[-\frac{(u-v)^2}{2B} \right]~\exp \left[-\frac{(u+v)^2}{2C} \right]\mathrm{d}u~\mathrm{d}v$$

We can easily see that:


Thus, we only need to find $I(B,C)$. Let's expand the squares in the exponentials:

$$I(B,C)=\int_0^1 \int_0^1 u~v~\exp (-\alpha ~u~v )~\exp \left[-\beta (u^2+v^2) \right]\mathrm{d}u~\mathrm{d}v$$

$$\alpha=\frac{1}{C}-\frac{1}{B}, \qquad \beta=\frac{1}{2C}+\frac{1}{2B}$$

Let's introduce a new integral:

$$J(\alpha,\beta)=\int_0^1 \int_0^1 \exp (-\alpha ~u~v )~\exp \left[-\beta (u^2+v^2) \right]\mathrm{d}u~\mathrm{d}v$$

$$I(B,C)=-\frac{\partial J(\alpha,\beta)}{\partial \alpha}$$

Now we notice that the function under the integral is symmetric with respect to the change $u \to v$, thus we can write:

$$J(\alpha,\beta)=2\int_0^1 \int_0^u \exp (-\alpha ~u~v )~\exp \left[-\beta (u^2+v^2) \right]\mathrm{d}u~\mathrm{d}v$$

Now we make a change of variable:

$$v=u t, \qquad dv=u~dt$$

$$J(\alpha,\beta)=2\int_0^1 \int_0^1 \exp (-\alpha ~u^2~t )~\exp \left[-\beta u^2(1+t^2) \right]u~\mathrm{d}u~\mathrm{d}t$$

Another change of variable will be:


$$J(\alpha,\beta)=\int_0^1 \int_0^1 \exp [-(\beta t^2+\alpha ~t+\beta)p ]\mathrm{d}p~\mathrm{d}t$$

The inner integral is elementary and we obtain a single integral:

$$J(\alpha,\beta)=\int_0^1 \frac{1-\exp [-(\beta t^2+\alpha ~t+\beta) ]}{\beta t^2+\alpha ~t+\beta}~\mathrm{d}t$$

$$=\frac{1}{\beta} \int_0^1 \frac{\mathrm{d}t}{(t+\gamma)^2 +1-\gamma^2}-e^{\beta (1-\gamma^2)} \frac{1}{\beta} \int_0^1 \frac{\exp [-\beta (t+\gamma)^2 ]}{(t+\gamma)^2 +1-\gamma^2}~\mathrm{d}t$$

$$\gamma=\frac{\alpha}{2 \beta}=\frac{B-C}{B+C}$$

The first part is an elementary integral (arctangent if $bc>0$).

The second part is related to the error function.

Let's get back to the initial integral and express it in the new form. Note that the exchange $B \to C$ is equivalent to $\alpha \to - \alpha$ and $\beta \to \beta$:


$$=2a^4\sqrt{\frac{2}{bc}} \left(-\frac{\partial J(\alpha,\beta)}{\partial \alpha}+\frac{\partial J(-\alpha,\beta)}{\partial \alpha} \right) =2a^4\sqrt{\frac{2}{bc}} \times$$

$$ \times \int_0^1 \left[ \frac{ 1-\left(\beta t^2 +\alpha t+\beta+1\right) e^{-(\beta t^2 +\alpha t+\beta)}}{(\beta t^2 +\alpha t+\beta)^2}-\frac{ 1-\left(\beta t^2 -\alpha t+\beta+1\right) e^{-(\beta t^2 -\alpha t+\beta)}}{(\beta t^2 -\alpha t+\beta)^2}~\right] t \mathrm{d}t $$

Numerically the first and last expressions are equal, which was checked with Mathematica for random values of $a,b,c$.


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