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What is the value of this nontrivial itegral:

$$\int_0^{+\infty} \left( \prod_{n = 1}^{+\infty} \cos \frac{x}{n}\right) \, \mbox d x$$

I don't know if there is nice closed answer with known constants.

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Some thoughts: By Viete's formula we have $\displaystyle \frac{\sin(x)}{x} = \cos\frac{x}{2} \cdot \cos\frac{x}{4} \cdot \cos\frac{x}{8} \cdots =\prod_{i=1}^{\infty} \cos\frac{x}{2^n}$. From here you just have to evaluate $\int_{0}^{+\infty} \frac{\sin{x}}{x} =\frac{\pi}{2}$. –  user9413 May 31 '12 at 12:16
    
I know this, but in denominator I have linear $n$ instead of exponantial $2^n$ –  old May 31 '12 at 12:24
    
Of course one can numerically evaluate the problem and thus obtain $0.785381$. I'm not sure if this is any otherwise known number... Maybe we should call it the user13763-number. –  Fabian May 31 '12 at 12:59
    
It will be cool if it were $\frac{\pi}{4}$ –  old May 31 '12 at 13:17
3  
Just to be sure, but I think if you are asking this question you are aware of ams.org/notices/201110/rtx111001410p.pdf (section "Limits of computation"). The similar integral is the first term in the expansion of $\frac{\pi}{8}$. This expansion is quite good --- the first term gives 42 correct digits. Follow the references there too, it might help. –  Yrogirg May 31 '12 at 15:56

1 Answer 1

Beginning of an answer. Use these: $$\begin{align} \cos x &= \prod_{k=0}^\infty \left(1-\frac{4x^2}{(2k+1)^2 \pi^2}\right) \\ \frac{\sin x}{x} &= \prod_{k=1}^\infty \left(1-\frac{x^2}{k^2 \pi^2}\right) \end{align}$$ so that $$\begin{align} f(x) &:= \prod_{n=1}^\infty \cos \frac{x}{n} = \prod_{n=1}^\infty \prod_{k=0}^\infty \left(1-\frac{4x^2}{(2k+1)^2n^2\pi^2}\right) \\ &= \prod_{k=0}^\infty \prod_{n=1}^\infty \left(1-\frac{4x^2}{(2k+1)^2n^2\pi^2}\right) = \prod_{k=0}^\infty \frac{\sin\frac{2x}{2k+1}}{\frac{2x}{2k+1}} . \end{align}$$ We have to check that the order can be reversed.

Now (at least for the first few $K$)$^*$ I get $$ \int_0^\infty \prod_{k=0}^K \frac{\sin\frac{2x}{2k+1}}{\frac{2x}{2k+1}}\,dx = \frac{\pi}{4} $$ exactly. If we can find the right limit theorem, perhaps also $$ \int_0^\infty f(x)\,dx = \int_0^\infty \prod_{k=0}^\infty \frac{\sin\frac{2x}{2k+1}}{\frac{2x}{2k+1}}\,dx = \frac{\pi}{4} $$

$^*$ added No, the answer $\pi/4$ is only true up to $K=6$, but fails for $7$ and up.

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This might be related. –  Fabian Jun 1 '12 at 8:18
    
@GEdgar:We know now that the result will be less than π/4. I wonder if a value can be found that the integral result is bigger than the value. Thus the result can be narrowed and be bordered with 2 values. Thanks a lot for your answer. –  Mathlover Jun 1 '12 at 13:41
    
@Fabian: Yes, exactly (4) and (5) of that paper. –  GEdgar Jun 1 '12 at 14:11
    
@Mathlover: Why do we know it is less than $\pi/4$? These integrands are not nonnegative, so we don't know they decrease with $K$...??? –  GEdgar Jun 1 '12 at 14:13
    
I just read the user13763's comment above. –  Mathlover Jun 1 '12 at 14:19

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