# Principal Value of $\iint_{-\infty}^{\infty}\text{Re} \Bigl( \frac{1}{\sqrt{(y_2+iy_1)^2-b^2}}\Bigr) e^{i(k_1y_1+k_2y_2)}\mathrm{d}y_1\mathrm{d}y_2$

Does the integral

\begin{align} I = \iint_{-\infty}^{\infty}Re \left( \frac{1}{\sqrt{(y_2+iy_1)^2-b^2}} \right ) e^{i(k_1y_1+k_2y_2)}\mathrm{d}y_1\mathrm{d}y_2 \end{align} with $b,y_1,y_2, k_1, k_2$ all real, have an analytical solution? I would also be interested in solutions involving generalised functions if the integral does not converge.

Using the fact that $Re(z) = \frac{z+\bar{z}}{2}$ and attempting to evaluate the $y_2$ integral first we have to solve an integral of the form

\begin{align} I = \int_{-\infty}^{\infty} \frac{\cos{k_2y_2}}{\sqrt{y_2^2+(i2y_1)y_2-y_1^2-b^2}} \mathrm{d}y_2 \end{align}

And consulting "Table of Integrals, Series, an Products" by Gradshteyn and Ryzhik the closest tabulated integral that comes close to helping me is a solution to

\begin{align} I = \int_{0}^{\infty} (x^2+2\beta x)^{\nu-1/2}\cos(ax) \mathrm{d}x \end{align}

which has an analytical solution in terms of bessel and gamma functions. However this is as far as I get because of the missing constant under the square root sign...

• Are you sure the integral is written correctly? It looks like we have a non-integrable singularity in $(y_1,y_2)=(0,b)$. – Jack D'Aurizio Feb 16 '15 at 19:10
• Thanks for your comment, yes it should be correct so I'm also interested in solutions involving generalised functions. Sorry if this was not clear. – Dipole Feb 16 '15 at 19:12
• How are we supposed to deal with the non-integrable singularity? Should we take the principal value? In such a case, what is the determination of the square root we are considering here? – Jack D'Aurizio Feb 16 '15 at 19:15
• You should said something along Jack D'Aurizio comment plus some additional comment about the square root branch cut. – Felix Marin Feb 16 '15 at 19:48
• @JackD'Aurizio Yes, exactly I would like to take the principal value and define the square root to have branch cuts from $y_2=b$ to $\infty$ and $y_2=-b$ to $-\infty$ does that make sense? – Dipole Feb 17 '15 at 15:57