# Can Mathematica calculate the triple integral $\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{(1+x^2+y^2+z^2)^2}$?

A recent post in Mathematics Stack Exchange claims that one can get from Mathematica the following result: $$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{(1+x^2+y^2+z^2)^2}=\frac{\pi^2}{32}.$$

However, I am not able to reproduce the result with Mathematica 11. With the code

Integrate[1/(1+x^2+y^2+z^2)^2,{x,0,1},{y,0,1},{z,0,1}]

I only got: With WolframAlpha, one has a numerical result: Question: Can Mathematica calculate the mentioned triple integral?

[Added later:] According to what the author later added in his post, he actually calculated the integral with Mathematica only numerically. Lele's answer below thus answers the question in the title: yes, the triple integral can be calculated, numerically. I am also curious to know though if Mathematica can return the result as $$\dfrac{\pi^2}{32}$$.

• Please post the code you used to attempt to do this integral. – bbgodfrey Sep 11 at 17:24
• @bbgodfrey: it is nothing but one line and it is in the screen shot. Fair enough. I have now included it in the post. Thanks for your comment. – Jack Sep 11 at 17:41
• Version 12 gives essentially the same answer, Integrate[ArcCot[Sqrt[2 + x^2]]/((1 + x^2)*Sqrt[2 + x^2]), {x, 0, 1}]. I have no idea why the question was down-voted. – bbgodfrey Sep 11 at 18:07

For numerical integrations, you should use NIntegrate instead of Integrate.

With

NIntegrate[1/(1 + x^2 + y^2 + z^2)^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}]


you obtain the same result as in WolframAlpha: • A symbolic answer appears to be desired. – bbgodfrey Sep 11 at 17:25
• Ok, the OP has changed the question now – Lele Sep 11 at 17:44
• @Lele: +1 Thanks for your answer and sorry for the ambiguity. I will try to find better way to edit my post. – Jack Sep 11 at 17:48