Monday, 16 September 2019

integration - show that intinfty0fracsin(ax)x(x2+b2)2dx=fracpi2b4(1fraceab(ab+2)2)



show that




0sin(ax)x(x2+b2)2dx=π2b4(1eab(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(0sinaxxdx0xsinaxdxx2+b2)1b20xsinaxdx(x2+b2)2




The first integral is well known. For any a>0:



0sinaxxdx=π2



The second, consider:



f(t)=0xsinaxtdxx2+b2dxL{f(t)}=0est0xsinaxtdxx2+b2dxdt=0xx2+b20estsinaxtdtdx=0ax2(x2+b2)(a2x2+s2)dx=π2(s+ab)



π2L1{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)20estsinaxtdtdx=0ax2(x2+b2)2(a2x2+s2)dx=aπ4b(s+ab)2



πa4bL1{1(s+ab)2}|t=1=aπ4beab



Therefore:



0sinaxdxx(x2+b2)2=π2b4(12+ab2eab)



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