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I am doing the following integral in Mathematica with $a>0$:

$$\iint e^{-\frac{(x_{1}+x_{2}-2b)^2}{4a}}dx_{1}dx_{2}$$

My code is

Integrate[
 Exp[-2 ((x + y)/2 - b)^2/(2*a)], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]
(* ∞ Sqrt[Sign[a]] *)

Although the integration is infinity, I wish to obtain an approximate close form of the integral as a function of a, b. Is this possible using Mathematica ? How to do this?

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  • $\begingroup$ You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is also useful for learning how to format your questions and answers. You may also find this this meta Q&A helpful $\endgroup$
    – Michael E2
    Feb 9, 2016 at 22:31
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    $\begingroup$ Just to be clear, you wish to approximate infinity as a function of two variables? $\endgroup$
    – Michael E2
    Feb 9, 2016 at 22:35

2 Answers 2

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What about computing a general $t$-bound integral:

expr = Integrate[Exp[-2 ((x + y)/2 - b)^2/(2*a)], {x, -t, t}, {y, -t, t}];

And then expanding in series around $t=\infty$:

Series[expr, {t, Infinity, 3}] // Normal // PowerExpand // FullSimplify

$\frac{a^2 \left(e^{-\frac{(b-t)^2}{a}}+e^{-\frac{(b+t)^2}{a}}\right)}{t^2}-4 a e^{-\frac{b^2}{a}}-4 \sqrt{\pi } \sqrt{a} b \text{erf}\left(\frac{b}{\sqrt{a}}\right)+4 \sqrt{\pi } \sqrt{a} t$

You can see that at this limit expression tends to infinity linearly in $t$ as

$4 \sqrt{\pi } \sqrt{a} t$

because other terms become smaller as $t$ goes to infinity. You better check this logic, I did not dwell on this. Independence of $b$ makes sense.

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  • $\begingroup$ thx, this is valuable to restrict within finite bounds t. $\endgroup$
    – lzstat
    Feb 10, 2016 at 15:17
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If you change the variables: y1=x1+x2; y2=x2 you get to another expression:

enter image description here

where all the integration limits are +/- infinities. The first integral is just equal to infinity, while the second is

    Integrate[Exp[-(y1 - 2 b)^2/(4 a)], {y1, -\[Infinity], \[Infinity]}, 
 Assumptions -> {a > 0, b > 0}]

(* 2 Sqrt[a] Sqrt[\[Pi]]  *)

I hope this helps. Have fun!

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  • $\begingroup$ thx You were right after change of variables, we could have a expression. $\endgroup$
    – lzstat
    Feb 10, 2016 at 15:18

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