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A friend and I have been looking at the crazy integral $$\iiiint \limits^{\infty}_{-\infty}\exp\left[-(x-t)^2-(x-h)^2-(y+t)^2-(y-h)^2-10\right]\mathrm{d}V$$ and can't come up with a decent method on how to obtain a solution. ($x,y,t,h$ are all variables, not constants) Fubini's theorem would let it split into 4 integrals, but those aren't the cleanest either. I can't think of another method that will work (of course u-sub/parts, etc).

Could residue theorem/contour integration work? Fourier integral? Any hints would be appreciated.

Edit -- changed $y-t$ to $y+t$

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    $\begingroup$ Use en.wikipedia.org/wiki/… $\endgroup$ – mlainz Apr 19 '16 at 20:56
  • $\begingroup$ You need to find the $A_{i,j}$ and the $B_i$. $\endgroup$ – mlainz Apr 19 '16 at 20:56
  • $\begingroup$ How is that used? I'm not sure where matrices work into this.. $\endgroup$ – galois Apr 19 '16 at 21:33
  • $\begingroup$ Also, I made a change, small but still there. $\endgroup$ – galois Apr 19 '16 at 21:36
  • $\begingroup$ Expand it as a polynomial. $A_{i,i}$ are the coefficients of $x_i^2$ $2 A_{i,j}$ are the coefficients of $x_i x_j$. The linear part has coefficients $B_i$. The constant part is not a problem. You can take it out of the integral. $\endgroup$ – mlainz Apr 19 '16 at 22:57
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According to http://m.wolframalpha.com/input/?i=%28x-t%29%5E2+%2B+%28x-h%29%5E2+%2B+%28y%2Bt%29%5E2+%2B+%28y-h%29%5E2&x=0&y=0

$$(x - t)^2 + (x - h)^2 + (y + t)^2 + (y - h)^2 =2 h^2 + 2 t^2 - 2 h x - 2 t x + 2 x^2 - 2 h y + 2 t y + 2 y^2$$

Therefore, if $\vec x = (x,y,t, h)$, then $$e^{-10} \int \exp{-\frac12 (\vec x \cdot A \cdot \vec x)},$$

Where $$A = \left( \begin{array}{cccc} 4 & 0 & -2 & -2 \\ 0 & 4 & 2 & -2 \\ -2 & 2 & 4 & 0 \\ -2 & -2 & 0 & 4 \\ \end{array} \right)$$

You can check that $A$ is positive definite. The result is $$e^{-10}\sqrt{\frac{π^4}{\det A}} = e^{-10} \frac{\pi^2}{8}$$

EDIT: The trick is that, if you have a quadratic form $q(\vec x)$ (a polynomial with every term of degree two $q(x_1,x_2,\ldots) = A_{1,1}x_1^2 + A_{1,2}x_1 x_2 + A_{2,2}x_2^2 + A_{1,3}x_1x_3 + \cdots$), and $q$ is positive definite ($q(\vec x)$ is always positive), then there is a linear change of variables $T$ that diagonalises $q$ (it converts it to $x_1^{\prime2}+x_2^{\prime2}+x_3^{\prime2}+\cdots x_n^{\prime2}$). You can prove that $dV = \frac{dV'}{\sqrt{\det{A}}}$. So

$$\int \exp{-q(\vec x)} dV = \int \exp{(-x_1^{\prime2}-x_2^{\prime2}-\cdots x_n^{\prime2})}\frac{dV'}{\sqrt{\det{A}}} =\\ \frac{1}{\sqrt{\det{A}}} \int \exp{(-x_1^{\prime2})} dx_1 \int \exp{(-x_2^{\prime2})} dx_2 \cdots \int \exp{(-x_n^{\prime2})} dx_n =\\ \frac{1}{\sqrt{\det{A}}} \left(\int \exp{(-t^2)} dt \right)^n, $$ which is the integral of the Gaussian function.

This integrals are actually very useful in probability theory when you want to study the distribution of correlated random variables.

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  • $\begingroup$ Interesting, I'm getting $e^{-10}\frac{\pi}{2}$, and Wolfram Alpha, too - goo.gl/uazyMV $\endgroup$ – galois Apr 20 '16 at 0:29
  • $\begingroup$ Sorry. I think I made a mistake when dividing by 2. But you got the point, right? $\endgroup$ – mlainz Apr 20 '16 at 2:04
  • $\begingroup$ So we can reduce a quadruple integral into a single integral by this method? That's neat $\endgroup$ – galois Apr 20 '16 at 2:37
  • $\begingroup$ Could you explain 2 things: how you found the entries of A in this case, and what you mean by $dV=\frac {dV'} {\sqrt {\det A }}$? $\endgroup$ – galois Apr 20 '16 at 14:39
  • $\begingroup$ The entries are the coefficients of the polynomial. There is a factor of two for the mixed terms $x_i x_j$, because you get both $a_{i,j}$ and $a_{j,i}$. $\endgroup$ – mlainz Apr 20 '16 at 17:24

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