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

 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?

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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


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.

  • $\begingroup$ thx, this is valuable to restrict within finite bounds t. $\endgroup$
    – lzstat
    Feb 10, 2016 at 15:17

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!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.