Reduction formula for Integral I am trying to find a reduction formula for $$\int\frac{\sin ^n(x)}{x} dx  .$$
Some numerical calculations seem to indicate a reduction formula should exist, but the usual tricks (break up the $\sin ^n(x)$ etc)  don't seem to help.
Any pointers/solutions would be great. 
 A: See that:
$$
\sin(nx)=\Im\{e^{inx}\}=\Im\{(\cos x+i\sin x)^n\}
$$
which is:
$$
(\cos x+i\sin x)^n=\sum_{k=0}^n \binom{n}{k}i^k\sin^k{x}\cos^{n-k}x.
$$
Let's consider first odd $n$'s:
$$
\sin{(2n+1)x}=\sum_{k=0}^n \binom {2n+1}{2k+1}(-1)^k\sin^{2k+1}x(1-\sin^{2}x)^{n-k}
$$
Therefore $\sin{(2n+1)x}$ can be written as ${\sum_{k=0}^{n}}{a_k\sin^{2k}x}$ for $a_k$ obtained from the previous relation. Using this the following recursive equation can be obtained relating $I_{2n+1}$ to other $I_{2k+1}$:
$$
I_{1}=\int_0^\infty\frac{\sin x}xdx=\int_{0}^\infty \frac{\sin(2n+1)x}{x}dx\\
=\int_{0}^\infty\frac{{\sum_{k=0}^{n}}{a_k\sin^{2k}x}}{x}=\sum_{k=0}^n a_kI_{2k+1}.
$$

Even $n$:
First define
$$
J_n=\int_0^\infty \frac{\cos^n x}{x}dx.
$$
Using a similar argument to above, we can write $\cos(2nx)=\sum_{k=0}^nb_k\cos^{2k}x$ and hence:
$$
J_1=\sum_{k=0}^nb_kJ_{2k}.
$$
Therefore all $J_{2k}$'s are covergent. Now see that:
$$
\int_0^\infty \frac{\sin^{2n}x}{x}dx=\int_0^\infty \frac{(1-\cos^{2}x)^n}{x}dx\\
=\int_{0}^\infty \frac 1xdx+\sum_{k=1}^n\binom{n}{k}(-1)^kJ_{2k},
$$
which is divergenct.
A: This is not an answer, but can lead you to the answer
$\text{Si}=\int\frac{\sin (x)}{x} \,dx;\;\text{Ci}=\int\frac{\cos (x)}{x} \,dx$
Some solutions of 
$$\int\frac{\sin ^n(x)}{x} \,dx$$
for $n=1..10$
$$
\begin{array}{l|l}
n & integral\\
\hline
 1 & \text{Si}(x) \\
 2 & \frac{\log (x)}{2}-\frac{\text{Ci}(2 x)}{2} \\
 3 & \frac{1}{4} (3 \text{Si}(x)-\text{Si}(3 x)) \\
 4 & \frac{1}{8} (-4 \text{Ci}(2 x)+\text{Ci}(4 x)+3 \log (x)) \\
 5 & \frac{1}{16} (10 \text{Si}(x)-5 \text{Si}(3 x)+\text{Si}(5 x)) \\
 6 & \frac{1}{32} (-15 \text{Ci}(2 x)+6 \text{Ci}(4 x)-\text{Ci}(6 x)+10 \log (x)) \\
 7 & \frac{1}{64} (35 \text{Si}(x)-21 \text{Si}(3 x)+7 \text{Si}(5 x)-\text{Si}(7 x)) \\
 8 & \frac{1}{128} (-56 \text{Ci}(2 x)+28 \text{Ci}(4 x)-8 \text{Ci}(6 x)+\text{Ci}(8 x)+35 \log (x)) \\
 9 & \frac{1}{256} (126 \text{Si}(x)-84 \text{Si}(3 x)+36 \text{Si}(5 x)-9 \text{Si}(7 x)+\text{Si}(9 x)) \\
 10 & \frac{1}{512} (-210 \text{Ci}(2 x)+120 \text{Ci}(4 x)-45 \text{Ci}(6 x)+10 \text{Ci}(8 x)-\text{Ci}(10 x)+126 \log (x)) \\
\end{array}
$$
