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Determine, with justification, the value of the integral $\int_0^1 \int_0^1 \int_0^1 \frac{1}{(1+x^2+y^2+z^2)^2} dxdydz$.

I tried converting this integral to cylindrical coordinates with $r = \sqrt{x^2 + y^2}$ ranging from $0$ to $\sqrt{2}$, $0\leq \theta \leq \pi/2, 0\leq z \leq 1,$ where $\theta $ is such that $x= r\cos\theta, y = r\sin\theta.$ However, this seems to lead to an incorrect result. Which bounds have I gotten wrong? Also, it seems that the integral over the unit cube equals twice the integral over the region defined by $0\leq z\leq 1, 0\leq x\leq 1, 0\leq y\leq x,$ but I'm not sure why. The result should be $\frac{\pi^2}{32},$ which is basically what WolframAlpha outputs.

Using spherical coordinates seems to make the integration more complicated due to the integration factor.

user3472
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    The general formula $$I(n)=\int_{[0,1]^{n-1}}\frac{1}{(1+{x_1}^2+{x_2}^2+…+{x_{n-1}}^2)^{n/2}}dx_1 … dx_{n-1}=\frac{(\sqrt\pi)^n}{2^n\Gamma\Big(\frac{n}{2}+1\Big)}$$ can be derived (solutions: https://math.stackexchange.com/questions/4268196/an-integral-over-n-dimensional-unit-cube/4268547#4268547 or https://mathoverflow.net/questions/319058/integrating-over-a-hypercube-not-a-hypersphere/319068#319068 ).

    For the case $n=4$ it gives: $$ I(4)=\int_{[0,1]^3}\frac{1}{(1+x^2+y^2+z^2)^2}dx,dy,dz=\frac{\pi^2}{32}$$

     – Svyatoslav Oct 21 '21 at 05:26
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    @Svyatoslav Nice. Thank you. – River Li Oct 21 '21 at 07:17

1 Answers1

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Using the identity ($q > 0$) $$\frac{1}{q^2} = \int_0^\infty t\mathrm{e}^{-qt} \,\mathrm{d} t,$$ we have \begin{align*} I &= \int_0^1\int_0^1 \int_0^1 \int_0^\infty t\mathrm{e}^{-(1 + x^2 + y^2 + z^2)t} \,\mathrm{d} t \,\mathrm{d}x \mathrm{d}y \mathrm{d}z\\ &= \int_0^\infty t\mathrm{e}^{-t} \left(\int_0^1 \mathrm{e}^{- x^2t}\mathrm{d} x\right)^3\mathrm{d}t\\ &= \int_0^\infty t\mathrm{e}^{-t} \left(\frac{1}{\sqrt{t}}\int_0^{\sqrt{t}} \mathrm{e}^{- u^2}\mathrm{d} u\right)^3\mathrm{d}t\\ &= \int_0^\infty \frac{1}{\sqrt{t}}\mathrm{e}^{-t} \left(\int_0^{\sqrt{t}} \mathrm{e}^{- u^2}\mathrm{d} u\right)^3\mathrm{d}t \\ &= \frac12\left(\int_0^{\sqrt{t}} \mathrm{e}^{- u^2}\mathrm{d} u\right)^4\Big\vert_0^\infty \\ &= \frac12 \left(\int_0^{\infty} \mathrm{e}^{- u^2}\mathrm{d} u\right)^4\\ &= \frac12\left(\frac12\sqrt{\pi}\right)^4\\ &= \frac{\pi^2}{32} \end{align*} where we have used $\int_{-\infty}^\infty \mathrm{e}^{-u^2}\mathrm{d} u = \sqrt{\pi}$, and $(\int_0^{\sqrt{t}} \mathrm{e}^{- u^2}\mathrm{d} u)' = \frac{1}{2\sqrt{t}}\mathrm{e}^{-t}$.

River Li
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