I've used f(z) = ln(z + ib) / (z² + a²) , b>0 And then contour integration by using upper side rectangular contour with residue at z = ia , and then equating real parts and we got it.
One can make life simpler by choosing b as the Feynman parameter. dI/db = 2b \int^{\infty}_{-\infty} dx i/(x^2+a^2)(x^2+b^2) = 2\pi /a 1/(b+a) by contour integration. Then I =( 2\pi/a )ln(b+a) + c. Setting b=0, we get J=\int^{\infty}_{-\infty} dx ln(x^2)/(x^2+a^2) = (2/pi /a )ln a . Thus, c = 0. So, I = ( 2\pi/a )ln(b+a).
I've used f(z) = ln(z + ib) / (z² + a²) , b>0
And then contour integration by using upper side rectangular contour with residue at z = ia , and then equating real parts and we got it.
One can make life simpler by choosing b as the Feynman parameter. dI/db = 2b \int^{\infty}_{-\infty} dx i/(x^2+a^2)(x^2+b^2) = 2\pi /a 1/(b+a) by contour integration. Then I =( 2\pi/a )ln(b+a) + c. Setting b=0, we get J=\int^{\infty}_{-\infty} dx ln(x^2)/(x^2+a^2) = (2/pi /a )ln a . Thus, c = 0. So, I = ( 2\pi/a )ln(b+a).
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