Strange integral

Calculus Level 5

∫ 0 ∞ ∏ k = 0 n s i n c ( x 2 k + 1 ) d x = π 2 \def\sinc{sinc\,} \large \int_{0}^{\infty} \prod\limits_{k=0}^{n}\sinc\left(\frac{x}{2k+1}\right)dx=\, \frac{\pi}{2}

Find the largest value of n n which satisfies the above equation.

• n n is a natural number.

• s i n c ( x ) = sin ⁡ x x . \def\sinc{sinc\,} \sinc (x)=\frac{\sin x}{x}.


The answer is 6.

Isaac Buckley by Isaac Buckley , 24, United Kingdom

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1 solution

Kazem Sepehrinia
Jul 20, 2015

Borwein integral! Why on earth should that be so?!!

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I have no idea Kazem... It blows my mind. I'm trying to get my puny brain to understand it. This isn't the only interesting Sinc product result either! There are a lot and from the papers I've read about it all this one has something to do with this fact:

∑ n = 0 6 1 2 n + 1 < 2 \large \sum\limits_{n=0}^{6}\frac1{2n+1}<2

But, ∑ n = 0 7 1 2 n + 1 > 2 \large \sum\limits_{n=0}^{7}\frac1{2n+1}>2

I don't know enough Fourier analysis to be sure though.

Here are the two best papers I found: Paper 1 and Paper 2

Isaac Buckley - 5 years, 10 months ago

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Isaac these are really strange! I was reading this one . I can't understand some points of it, but the proof and result are indescribably beautiful. Thanks for sharing this.

Kazem Sepehrinia - 5 years, 10 months ago

I was actually looking for a proof as well. It's frustrating that it's hard to find proofs for beautiful identities like these.

To make the question more complicated, add a 2 cos ⁡ ( x ) 2\cos(x) in front of the product notation.

Pi Han Goh - 5 years, 10 months ago

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I think if you put ∑ n = 0 m 2 cos ⁡ ( ( 2 n + 1 ) x ) ) \sum\limits_{n=0}^{m} 2\cos\left((2n+1)x)\right) it works.

I think the "paper 2" I sited suggests this result:

\DeclareMathOperator ∗ s i n c s i n c ∫ 0 ∞ ( 2 cos ⁡ ( x ) + 2 cos ⁡ ( 3 x ) + 2 cos ⁡ ( 5 x ) ) ∏ k = 0 n \sinc ( x 2 k + 1 ) d x = π 2 \DeclareMathOperator*{sinc}{sinc\,} \large \int_{0}^{\infty} \left(2\cos(x)+2\cos(3x)+2\cos(5x)\right) \prod\limits_{k=0}^{n}\sinc\left(\frac{x}{2k+1}\right)dx=\, \frac{π}{2}

Works for all n = 1 , 2 , 3 , . . . , 168802 n=1,2,3,...,168802 .

Why 168802? Well I think it's because

∑ n = 0 168802 1 2 n + 1 = 6.99999... < 7 \large \sum\limits_{n=0}^{168802}\frac1{2n+1}=6.99999...<7

I really am clueless... Hopefully someone could shed some light to this beautiful mystery.

"I think"

Isaac Buckley - 5 years, 10 months ago

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@Isaac Buckley – Oh ahah, I was actually referring to the one at the bottom on this page . But yours is clearly more interesting!

Pi Han Goh - 5 years, 10 months ago

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@Pi Han Goh – Why do i feel that Gamma function will be useful for his integral :P

akash omble - 5 years, 10 months ago

It's really interesting!! It just shows pure beauty of mathematics! With an easy pure application of Fourier Transform, Borwein just made me his fan.

Thanks a lot for the problem! @Isaac Buckley Hope to see more of such interesting problems from you! :)

Kartik Sharma - 5 years, 10 months ago

Relevant question .

Pi Han Goh - 5 years, 5 months ago

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