0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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