How to find $\int_0^\infty \prod_{k=1}^n \frac{\sin \frac{x}{2k-1}}{\frac{x}{2k-1}}\mathrm dx$ 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..
 A: 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 $I_8$.
