I am trying to solve this integral and limit;

$$\lim_{n\to \infty}{\int_0^n\int_0^n\int_0^n\int_0^n\sqrt{\left({e-b\over n}\right)^2+\left({c-a\over n}\right)^2}de\ dc\ db\ da\over n^4}$$

I tried using wolfram alpha to calculate it, using;

lim ((integral 0 to n of (integral 0 to n of (integral 0 to n of (integral 0 to n of (sqrt(((e-b)/n)^2+((c-a)/n)^2)) de) dc) db) da)/n^4) as n->infinity

But wolfram alpha said it didn't understand the query - I believe because it is too long, as removing the limit calculation means it understands it.

Does anyone know how I can calculate this either by hand or using a program? I am interested in both. I have no idea how to solve it by hand.

  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Jul 8 '16 at 7:13
  • $\begingroup$ That is equal to: $$ \iint_{\left(0,1\right)^{2}}\sqrt{\left(e - b\right)^{2} + \left(c - a\right)^{2}}\, \mathrm{d}e\,\mathrm{d}c\,\mathrm{d}b\,\mathrm{d}a = {1 \over 15}\left[2 + \sqrt{2} + 5\ln\left(1 + \sqrt{2}\right)\right] \approx 0.5214 $$ I just evaluated overhere => math.stackexchange.com/a/1852733/85343 $\endgroup$ – Felix Marin Jul 8 '16 at 7:38
  • $\begingroup$ That problem you linked was the one I was trying to solve, funnily enough $\endgroup$ – Super Hacker Jul 16 '16 at 16:36

Well this is the wolfram query. You just need to add an extra limit to it. It exceeds standard computation time and I dont have access to Wolfram Alpha Pro, so I cant proceed further. If you have access, it will help.

(int_0^n int_0^n int_0^n int_o^n sqrt{((e-b)/n)^2+((c-a)/n)^2} db da dc de)/n^4


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