How to find $\int_0^\infty \prod_{k=1}^n \frac{\sin \frac{x}{2k-1}}{\frac{x}{2k-1}}\mathrm dx$

Solution 1:

Here is a partial answer, perhaps someone can fill in a bit more. Throughout the answer, $a_1,a_2,\ldots$ will be positive constants. We use the result $$\int_0^\infty \frac{\sin ax}{x}\,dx =\frac{\pi}{2}{\mathop{\rm sgn}}(a) =\cases{{\textstyle\frac{\pi}{2}}&if $a>0$\cr 0&if $a=0$\cr {\textstyle-\frac{\pi}{2}}&if $a<0$\cr}$$ together with trig identities and reversing a double integral to obtain $$\eqalign{\int_0^\infty\frac{\sin a_1x}{x}\frac{\sin a_2x}{x}\,dx &=\frac{1}{2} \int_0^\infty\frac{\cos(a_1-a_2)x-\cos(a_1+a_2)x}{x^2}\,dx\cr &=\frac{1}{2} \int_0^\infty\int_{a_1-a_2}^{a_1+a_2}\frac{\sin yx}{x}\,dy\,dx\cr &=\frac{\pi}{4} \int_{a_1-a_2}^{a_1+a_2}{\mathop{\rm sgn}}(y)\,dy\ .\cr}$$ If $a_2<a_1$ then the final integral involves only positive values of $y$, so we can write ${\mathop{\rm sgn}}(y)=1$ and we have $$\int_0^\infty\frac{\sin a_1x}{x}\frac{\sin a_2x}{x}\,dx =\frac{\pi}{2}a_2\ ;$$ taking $a_1=1$, $a_2=\frac{1}{3}$ and dividing by $a_1a_2$ gives your $I_2$. If on the other hand $a_2>a_1$ then we get $$\int_0^\infty\frac{\sin a_1x}{x}\frac{\sin a_2x}{x}\,dx =\frac{\pi}{2}a_1\ ;$$ the two results can be summed up as $$\int_0^\infty\frac{\sin a_1x}{x}\frac{\sin a_2x}{x}\,dx =\frac{\pi}{2}\min(a_1,a_2)\ ,$$

Next, a similar calculation gives $$\eqalign{\int_0^\infty\frac{\sin a_1x}{x}\frac{\sin a_2x}{x} \frac{\sin a_3x}{x}\,dx &=\int_0^\infty\frac{\sin a_1x}{x}\frac{1}{2} \int_{a_2-a_3}^{a_2+a_3}\frac{\sin yx}{x}\,dx\,dy\cr &=\frac{1}{2}\int_{a_2-a_3}^{a_2+a_3} \int_0^\infty\frac{\sin a_1x}{x}\frac{\sin yx}{x}\,dx\,dy\cr &=\frac{\pi}{4}\int_{a_2-a_3}^{a_2+a_3}\min(a_1,y)\,dy\ .\cr}$$ If $a_1-a_2-a_3>0$ then $\min(a_1,y)=y$ throughout the interval of integration and we have $$\int_0^\infty\frac{\sin a_1x}{x}\frac{\sin a_2x}{x} \frac{\sin a_3x}{x}\,dx =\frac{\pi}{8}\int_{a_2-a_3}^{a_2+a_3}2y\,dy=\frac{\pi}{2}a_2a_3\ ,$$ and once again taking suitable $a_1,a_2,a_3$ gives your $I_3$. We can use exactly the same ideas to prove by induction the following result.

Lemma. Suppose that $a_1>a_2>\cdots>a_n>0$ and $a_1-a_2-a_3-\cdots-a_n>0$. Then $$\int_0^\infty\frac{\sin a_1x}{x}\cdots\frac{\sin a_nx}{x}\,dx =\frac{\pi}{2}a_2a_3\cdots a_n\ .$$

Since $$1-\frac{1}{3}-\frac{1}{5}-\cdots-\frac{1}{13}>0\ ,$$ this proves your results for $I_1,\ldots,I_7$. However, $$1-\frac{1}{3}-\frac{1}{5}-\cdots-\frac{1}{13}-\frac{1}{15}$$ is negative, and so the lemma does not apply to $I_8$. It remains to determine whether or not something like these ideas can be used to evaluate $I_8$.