show that
∫∞0sin(ax)x(x2+b2)2dx=π2b4(1−e−ab(ab+2)2)
for a,b>0
I would like someone solve it using contour but also I would to see different solution using different way to solve it
is there any help
thanks for all
Answer
∫∞0sinaxdxx(x2+b2)2=1b4(∫∞0sinaxxdx−∫∞0xsinaxdxx2+b2)−1b2∫∞0xsinaxdx(x2+b2)2
The first integral is well known. For any a>0:
∫∞0sinaxxdx=π2
The second, consider:
f(t)=∫∞0xsinaxtdxx2+b2dx⇒L{f(t)}=∫∞0e−st∫∞0xsinaxtdxx2+b2dxdt=∫∞0xx2+b2∫∞0e−stsinaxtdtdx=∫∞0ax2(x2+b2)(a2x2+s2)dx=π2(s+ab)
π2⋅L−1{1s+ab}|t=1=π2eab
The third, using the same parameter (call the function g(t) now) one obtains:
L{g(t)}=∫∞0x(x2+b2)2∫∞0e−stsinaxtdtdx=∫∞0ax2(x2+b2)2(a2x2+s2)dx=aπ4b(s+ab)2
πa4b⋅L−1{1(s+ab)2}|t=1=aπ4beab
Therefore:
∫∞0sinaxdxx(x2+b2)2=π2b4(1−2+ab2eab)
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