# A multiple integral question II

We know from the previous post that $$\lim_{n\to\infty}\underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{n \text{ times}}\frac{1}{(x_1\cdot x_2\cdots x_n)^2+1} \mathrm{d}x_1\cdot\mathrm{d}x_2\cdots\mathrm{d}x_n=1$$

and now I wonder what is $f(n)$ such that

$$\lim_{n\to\infty}f(n)\left(1-\underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{n \text{ times}}\frac{1}{(x_1\cdot x_2\cdots x_n)^2+1} \mathrm{d}x_1\cdot\mathrm{d}x_2\cdots\mathrm{d}x_n\right)$$ is finite and then I'd like to find out its precise limit.

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You just make this stuff up, don't you? Some people cook, some crochet, some people have dogs, you sit around and make up problems. Do you get much sunshine? Also, you are evidently more than one person, and exist in separate bodies. You could go out bicycling together. – Will Jagy Feb 23 '13 at 7:16
Have you tried folk dancing? – Will Jagy Feb 23 '13 at 7:18
Now that I think of it, I went to a conference in 2010. An eleven year old girl also gave a talk, a prodigy. I suggested to her mother (attending as chaperone) that soccer might be a nice idea. However, the girl claimed that there was nothing more stupid than running around chasing a ball. Then I said, well maybe something solo, surely there is some sort of traditional dancing for girls (the parents are from India). The mother actually told me "Yes, but she sucks at it." So maybe you can keep on with the mathematics problems. – Will Jagy Feb 23 '13 at 7:23
@WillJagy: hehe, interesting. – Chris's sis the artist Feb 23 '13 at 7:28
@WillJagy: Anyway, thank you for sharing your thoughts. There are many other nice activities beside doing math and we shouldn't miss them. I understand your point and you're right. – Chris's sis the artist Feb 23 '13 at 8:23

Not sure about this but I'll write it anyway. Writing $\frac{1}{(x_1x_2..x_n)^2+1}$ as $\sum_{k=0}^\infty(-1)^k (x_1x_2..x_n)^{2k}$

We get,

$g(n)=\int_0^1\int_0^1\int_0^1....\int_0^1\frac{1}{(x_1x_2..x_n)^2+1}dx_1dx_2...dx_n$$=\int_0^1\int_0^1\int_0^1....\int_0^1\sum_{k=0}^\infty(-1)^k (x_1x_2..x_n)^{2k}dx_1dx_2...dx_n=\sum_{k=0}^\infty(-1)^k\frac{1}{(2k+1)^n}=\beta(n)$

(The Dirichlet beta function)

Note that $g(n)=1-3^{-n}+o(3^{-n})$

$f(n)=\frac{1}{1-g(n)}=3^n+o(3^n)$ is the function you want. Taking $f(n)=3^n$

$1\leq\lim_{n\to \infty} 3^n(1-g(n))\leq1-(\frac{3}{5})^n$

and by the squeeze theorem, $\lim{n\to\infty3^n(1-g(n))=1}$

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I like your work so far. Thanks. (+1) – Chris's sis the artist Feb 23 '13 at 7:50
Wolfram gives the answer in terms of the generalized zeta function, but I don't know if that's helpful. – Ishan Banerjee Feb 23 '13 at 8:10
I think the function is $\Im(Li_n(i))$ – Ishan Banerjee Feb 24 '13 at 5:32
@Chris'ssisterandpals Is this enough or is this non-rigorous or incomplete? – Ishan Banerjee Feb 25 '13 at 13:13
I just realized that something missed in my initial question and my expectations were a bit different. Anyway, I let it this way. Thanks for your work! – Chris's sis the artist Feb 25 '13 at 13:24