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I am trying to calculate the integral $$ I_n=\int \limits_0^\infty \prod_{k=1}^n \frac{\sin \frac{x}{2k-1}}{\frac{x}{2k-1}}\mathrm dx. $$ (I have literature on this, if people want). Note, we can write the amazing sequence $\{I_1,I_2,I_3,I_4,I_5,I_6,I_7\}$ as $$ \bigg\{\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2}\bigg\}. $$ BUT $I_8\neq \pi/2$, how can we derive this same result for $n=1,2,\ldots,7$? And why does it deviate at $I_8$? Thanks, in integral form this sequence is represented by $$ \frac{\pi}{2}=I_1=\int\limits_0^\infty \frac{\sin x}{x}\mathrm dx=I_2=\int \limits_0^\infty \frac{\sin x}{x}\frac{\sin \frac{x}{3}}{\frac{x}{3}}\mathrm dx=I_3=\int\limits_0^\infty \frac{\sin x}{x}\frac{\sin \frac{x}{3}}{\frac{x}{3}}\frac{\sin \frac{x}{5}}{\frac{x}{5}}\mathrm dx=\cdots $$ HOWEVER, this fails for $I_8$. The strange result for $I_8$ is given by $$ I_8= \frac{467807924713440738696537864469}{ 935615849440640907310521750000}\pi\approx \frac{\pi}{2}-2.31\cdot 10^{-11} $$

Note we can calculate $I_1$ by using integration wrt parameter and first considering the damped Sine- integral \begin{equation} \eta(\lambda)=\int_{0}^\infty e^{-\lambda x}\frac{\sin x}{x}\mathrm dx. \end{equation} We now wish to calculate the Dirichlet integral $I_1$ using calculus and $\eta(\lambda)$, \begin{equation} I_1=\int_{0}^\infty \frac{\sin x}{x}\mathrm dx. \end{equation} by differentiating $\eta(\lambda)$. We start by differentiating this to obtain $$ \eta'(\lambda)=\frac{d}{d\lambda} \int_{0}^{\infty} e^{-\lambda x}\frac{\sin x}{x}\mathrm dx =\int_{0}^\infty \frac{\partial}{\partial \lambda} e^{-\lambda x}\frac{\sin x}{x}\mathrm dx=-\int_{0}^\infty e^{-\lambda x}{\sin x}\ \mathrm dx. $$ Note, that passing the differentiation outside of the integral inside the integral is allowed since the integral is a continuous function of x and $\lambda$ for x$\in(-\infty,\infty)$ and $\lambda \in (0,\infty)$. We can easily integrate this by writing the sine function as the imaginary part of an exponential, that is $$ -\int_{0}^\infty e^{-\lambda x}{\sin x}\ \mathrm dx=-\Im\bigg[-\int_{0}^\infty e^{-\lambda x} e^{ix}\mathrm dx\bigg]=-\Im \bigg[-\int_{0}^\infty e^{-x(\lambda-i)}\mathrm dx\bigg]=-\Im{\frac{1}{\lambda-i}}=-\frac{1}{\lambda^2+1}, $$where I integrated the exponential using analysis rules and next used $$ -\Im\bigg [\frac{1}{\lambda-i}\bigg]=-\Im \bigg[\frac{1}{\lambda-i}\cdot \frac{\lambda+i}{\lambda+i}\bigg]=-\frac{1}{\lambda^2+1}. $$ Thus we can see that \begin{equation} \eta'(\lambda)= -\frac{1}{\lambda^2+1}. \end{equation} Now we need to use integrate this relation carefully. We do this by writing $$ \int_{\lambda}^{\infty}\frac{\mathrm d\eta}{\mathrm d\xi}\mathrm d\xi=\eta(\infty)-\eta(\lambda)=-\eta(\lambda) $$ since $\eta(\infty)=0$. We can now use this and the result above to give $$ -\eta(\lambda)=\int_{\lambda}^{\infty} \eta'(\xi)\mathrm d\xi=\int_{\lambda}^{\infty} -\frac{1}{\xi ^2 +1}\mathrm d\xi=-(\arctan{\infty}-\arctan{\lambda})=-\frac{\pi}{2}+\arctan{\lambda}, $$ thus we can easily see $$ \eta(\lambda)= \frac{\pi}{2}-\arctan{\lambda}. $$ We set $\lambda =0$ and obtain the desired result \begin{equation} \eta(\lambda=0)=I_1= \frac{\pi}{2}=\int_{0}^{\infty} \frac{\sin x}{x}\mathrm dx. \end{equation} But how to generalize this for $I_n$? Thanks a lot..

Ali Caglayan
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Jeff Faraci
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1 Answers1

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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 $I_8$.

David
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