How to evaluate L(m):=∫∞0(sin(x)x)mdx?
I am familiar with the case m=1, but what about the general one?
Answer
I recently explained this issue to a student in the following way:
Let f(x)=sinm(x) where m>1. Integration by parts m−1 times gives the formula
∫∞0f(x)xmdx=1(m−1)!∫∞0f(m−1)(x)xdx.
Now we have the formulae
\begin{align} \sin^{2n}(x)&=\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}(-1)^{n-k}\binom{2n}{k}\cos((2n-2k)x)+\frac{1}{2^{2n}}\binom{2n}{n} \\ \sin^{2n-1}(x)&=\frac{1}{2^{2n-2}}\sum_{k=0}^{n-1}(-1)^{n-k-1}\binom{2n-1}{k}\sin((2n-2k-1)x). \end{align}
Differentiate the former 2n-1 times and the latter 2n-2 times to obtain with the very first equation
\begin{align} L(2n)&=\frac{\pi}{2^{2n}(2n-1)!}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n}{k}(2n-2k)^{2n-1} \\ L(2n-1)&=\frac{\pi}{2^{2n-1}(2n-2)!}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n-1}{k}(2n-2k-1)^{2n-2}, \end{align}
since
\int_0^{\infty}\frac{\sin((2n-2k)x)}{x}dx=\int_0^{\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2}.
This yields in a campact shape
L(m)=\frac{\pi}{2^{m}(m-1)!}\sum_{k=0}^{\left \lfloor{\frac{m}{2}}\right \rfloor }(-1)^{k}\binom{m}{k}(m-2k)^{m-1}.
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