I wanted to find the asymptotic form of


for $M\rightarrow\infty$. I tried

AsymptoticIntegrate[Exp[M (x - 1/2)^2 + M (y - 1/2)^2], 
   {x, 0, 1}, {y,0,1}, {M, Infinity, 1}]

but it doesn't work. Apparently AsymptoticIntegrate cannot be used with multi-dimensional integrals? Is there a workaround, or am I doing something wrong?

  • $\begingroup$ Your integral Can be written as product of two onedimensional integrals, for each of Thema you can apply AsmptoticIngegrate grate $\endgroup$ – Ulrich Neumann Sep 14 '18 at 16:00
  • $\begingroup$ @UlrichNeumann Thanks, I know. This is just an example. $\endgroup$ – becko Sep 14 '18 at 16:43

I think docs page shows clearly that it is for 1D integrals and in your example you are misusing syntax specification for 2nd item here:


For workaround you can simply compute this, these are easy known formulas:

f[M_]=Integrate[Exp[M (x - 1/2)^2 + M (y - 1/2)^2], {x, 0, 1}, {y, 0, 1}]

$$\frac{\pi \text{erfi}\left(\frac{\sqrt{M}}{2}\right)^2}{M}$$

and then

Limit[f[M], M -> Infinity]

Out[]= Infinity

which is quite obvious. Generally for things that you cannot really integrate, you can exchange the order of Taylor series and integration to often get nice results, for example:

  Series[Sin[y/x Tan[a x/y]], {a, 0, 5}], 
{x, 1, 2}, {y, 1, 2}]

$$\frac{11 a^5}{200}+\frac{2 a^3}{9}+a$$

| improve this answer | |
  • $\begingroup$ I'd expect in future versions AsymptoticIntegrate be able to handle at least some simple multi-dimensional integrals like this one. Also note that I don't want the limit of f(M) as $M\rightarrow\infty$. But I could do Series[f[M], {M,Infinity, 1}] to get what I want. Thanks. $\endgroup$ – becko Sep 13 '18 at 23:00

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