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
sin2n(x)=122n−1n−1∑k=0(−1)n−k(2nk)cos((2n−2k)x)+122n(2nn)sin2n−1(x)=122n−2n−1∑k=0(−1)n−k−1(2n−1k)sin((2n−2k−1)x).
Differentiate the former 2n−1 times and the latter 2n−2 times to obtain with the very first equation
L(2n)=π22n(2n−1)!n−1∑k=0(−1)k(2nk)(2n−2k)2n−1L(2n−1)=π22n−1(2n−2)!n−1∑k=0(−1)k(2n−1k)(2n−2k−1)2n−2,
since
∫∞0sin((2n−2k)x)xdx=∫∞0sin(x)xdx=π2.
This yields in a campact shape
L(m)=π2m(m−1)!⌊m2⌋∑k=0(−1)k(mk)(m−2k)m−1.
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