I am looking for a compact form solution for this integral:
$$\psi (\nu) = \int_0^{\infty}\frac{e^{i \nu x}}{(1+a\,x)^2} dx,$$
where $i=\sqrt{-1}$, and $a>0$ is a constant and $\nu$ is Real . I searched the table of integrals book but didn't find what I am looking for. Does this mean that I need to find this integration numerically?
Thanks in advance
Best Answer
Integration by parts using $$u={{e}^{i\nu x}}and\ dv=\frac{1}{{{\left( 1+ax \right)}^{2}}}$$ yields:
$$I=\int_{0}^{\infty }{\frac{{{e}^{i\nu x}}}{{{\left( 1+ax \right)}^{2}}}dx}=\left. \frac{-{{e}^{i\nu x}}}{a\left( 1+ax \right)} \right|_{0}^{\infty }+\frac{i\nu }{a}\int_{0}^{\infty }{\frac{{{e}^{i\nu x}}}{\left( 1+ax \right)}dx}=\frac{-1}{a}+\frac{i\nu }{a}\int_{0}^{\infty }{\frac{{{e}^{i\nu x}}}{\left( 1+ax \right)}dx}$$
Now use $${{e}^{i\nu x}}=\ \cos \left( \nu x \right)+i\sin \left( \nu x \right)$$
$$I=\frac{-1}{a}+\frac{i\nu }{a}\int_{0}^{\infty }{\frac{\cos \left( \nu x \right)}{\left( 1+ax \right)}dx}-\frac{\nu }{a}\int_{0}^{\infty }{\frac{\sin \left( \nu x \right)}{\left( 1+ax \right)}dx}\ =\frac{-1}{a}+\frac{i\nu }{a}{{I}_{1}}-\frac{1}{a}{{I}_{2}}$$
Where these two integrals can be found in terms of cosine integral and sine integral.