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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.

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    $\begingroup$ Can you please indicate which numerical calculations seem to indicate a reduction formula should exist? $\endgroup$ – uniquesolution Aug 31 '17 at 6:44
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    $\begingroup$ Abusing notation a bit, integrating from 0 to $\infty$ yields a sequence of values (as $n$ ranges) $\lbrace \pi/2,\infty, \pi/4,\infty, 3\pi/16 \dots \rbrace $ . So it looks like the formula should relate $I_n$ to $I_{n-2}.$ $\endgroup$ – user434180 Aug 31 '17 at 6:46
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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.

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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 $n=1..10$ $$ \begin{array}{l|l} n & integral\\ \hline 1 & \text{Si}(x) \\ 2 & \frac{\log (x)}{2}-\frac{\text{Ci}(2 x)}{2} \\ 3 & \frac{1}{4} (3 \text{Si}(x)-\text{Si}(3 x)) \\ 4 & \frac{1}{8} (-4 \text{Ci}(2 x)+\text{Ci}(4 x)+3 \log (x)) \\ 5 & \frac{1}{16} (10 \text{Si}(x)-5 \text{Si}(3 x)+\text{Si}(5 x)) \\ 6 & \frac{1}{32} (-15 \text{Ci}(2 x)+6 \text{Ci}(4 x)-\text{Ci}(6 x)+10 \log (x)) \\ 7 & \frac{1}{64} (35 \text{Si}(x)-21 \text{Si}(3 x)+7 \text{Si}(5 x)-\text{Si}(7 x)) \\ 8 & \frac{1}{128} (-56 \text{Ci}(2 x)+28 \text{Ci}(4 x)-8 \text{Ci}(6 x)+\text{Ci}(8 x)+35 \log (x)) \\ 9 & \frac{1}{256} (126 \text{Si}(x)-84 \text{Si}(3 x)+36 \text{Si}(5 x)-9 \text{Si}(7 x)+\text{Si}(9 x)) \\ 10 & \frac{1}{512} (-210 \text{Ci}(2 x)+120 \text{Ci}(4 x)-45 \text{Ci}(6 x)+10 \text{Ci}(8 x)-\text{Ci}(10 x)+126 \log (x)) \\ \end{array} $$

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