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How should I find a value of these integrals: $$ A:=\int_{0}^{\infty}\frac{\sin(x)}{(x+1)^{2-\nu}x^{\nu}}dx \quad\text{and}\quad B:=\int_{0}^{\infty}\frac{\sin(x)}{(x+1)^{1-\nu}x^{\nu}}dx, $$ where $\nu\in(0,1)$.

Because I asked about a problem as having resembled this before (here), I have tried to solve by the same argument, i.e., using the Laplace transform but I haven't found the formula for the Laplace transform. I also have attacked by the method via the complex integral (which appears in the Fourier transform) but it didn't work well. I guess that Bessel functions appear as the solution. However, I don't know how a solution (with Bessel functions) is provided if I do it.

I'm glad if you give a solution as possible in detail.

Thank you.

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  • According to Mathematica the MeijerG is involved, so I'd start looking in the tables for integral representations for this function. – mickep Sep 24 '15 at 08:48
  • Thank you for comment. How is the form of solution by Mathematica? – user Sep 24 '15 at 09:05
  • (2^-n MeijerG[{{n/2, (1 + n)/2}, {}}, {{1/2, 1/2, 1}, {0}}, 1/ 4])/(Sqrt[\[Pi]] Gamma[2 - n]) for the first one, and (2^(-1 - n) MeijerG[{{n/2, (1 + n)/2}, {}}, {{0, 1/2, 1/2}, {0}}, 1/ 4])/(Sqrt[\[Pi]] Gamma[1 - n]) for the second one. Here, $n=\nu$. – mickep Sep 24 '15 at 09:09
  • In order for Bessel functions to appear, we must have some “nice” values of $\nu~($such as integers or half-integers$)$, and even then it is not guaranteed; and if even that fails to hold, then $($ generalized $)$ hypergeometric functions are unavoidable. – Lucian Sep 24 '15 at 13:25

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