Integral $\lim_{\epsilon\to 0^+} \int_{\Lambda/s}^\Lambda \int_{-\infty}^\infty \frac{i}{y^2-x^2+i\epsilon} dydx = \pi\log s$ I am stucked with the following integral
$$\lim_{\epsilon\to 0^+} \int_{\Lambda/s}^\Lambda \int_{-\infty}^\infty \frac{i}{y^2-x^2+i\epsilon} dydx = \pi\log s.$$
Here $s,\Lambda$ are positive constants. This integral is motivated from physics.
How can I evaluate this integral?
 A: Consider first
$$\int_{-\infty}^\infty \frac{i}{y^2-x^2+i\epsilon} dy$$
for fixed $x$. We will use the identity
$$\frac{1}{f+i\epsilon}=\mathrm{PV}\frac{1}{f}-i\pi\delta(f)\,,$$
where PV denotes the principal value.
Noting that
$$
\int_{-\infty}^{+\infty} \mathrm{PV}\frac{1}{y^2-x^2}dy=
\frac{1}{2x}\int_{-\infty}^{+\infty}  \mathrm{PV}\left(\frac{1}{y-x}-\frac{1}{y+x}\right)dy=0
$$
and
$$
\delta(y^2-x^2)dy=\frac{1}{2x}\left(
\delta(y-x)+\delta(y+x)
\right)dy\,,
$$
we then have
$$\int_{-\infty}^\infty \frac{i}{y^2-x^2+i\epsilon} dy=\frac{\pi}{x}\,.$$
The integral over $x$ then yields
$$\pi\int_{\Lambda/s}^{\Lambda}\frac{dx}{x} = \pi\log s\,.
$$
A: Write
$$ \frac{i}{y^2 - x^2 + i\epsilon} = \frac{i}{2\sqrt{x^2 - i\epsilon}} \left( \frac{1}{y - \sqrt{x^2 - i\epsilon}} - \frac{1}{y + \sqrt{x^2 - i\epsilon}} \right), $$
where $\sqrt{\,\cdot\,}$ is the principal square root. Also, note that for $a, b \in \mathbb{R}$ with $ b \neq 0$,
$$ \int_{-\infty}^{\infty} \frac{\mathrm{d}y}{y - (a+ib)}
\stackrel{(y \mapsto y+a)}= \int_{-\infty}^{\infty} \frac{\mathrm{d}y}{y - ib}
= \int_{-\infty}^{\infty} \frac{y+ib}{y^2 + b^2} \, \mathrm{d}y
= i \pi \operatorname{sign}(b). $$
Together with the fact that $\operatorname{Im}(\sqrt{x^2-i\epsilon}) < 0$, this gives
$$ \int_{-\infty}^{\infty} \frac{i}{y^2 - x^2 + i\epsilon} \, \mathrm{d}y
= \frac{i}{2\sqrt{x^2 - i\epsilon}} \bigl((-\pi i) - (\pi i)\bigr) = \frac{\pi}{\sqrt{x^2 - i\epsilon}}. $$
Finally, by the Dominated Convergence Theorem,
$$ \lim_{\epsilon \to 0^+} \int_{\Lambda/s}^{\Lambda} \int_{-\infty}^{\infty} \frac{i}{y^2 - x^2 + i\epsilon} \, \mathrm{d}y\mathrm{d}x
= \lim_{\epsilon \to 0^+} \int_{\Lambda/s}^{\Lambda} \frac{\pi}{\sqrt{x^2 - i\epsilon}} \, \mathrm{d}x
= \int_{\Lambda/s}^{\Lambda} \frac{\pi}{x} \, \mathrm{d}x
= \pi \log s. $$
A: First lets caculate our integral as a contour integral
$\oint_C \frac{-i}{z^2-x^2+i\epsilon}dz\\
=\oint_C \frac{-i}{(y+iu)^2-x^2+i\epsilon}d(y+iu)\\
=\int_{\{y\in(R,-R),u=0\}}\frac{-i}{(y+iu)^2-x^2+i\epsilon}d(y+iu)+\int_{\{y+íu=Re^{i\theta},\theta\in(-\pi/2,\pi/2)\}} \frac{-i}{(y+iu)^2-x^2+i\epsilon}d(y+iu)\\
=\int_R^{-R}\frac{-i}{y^2-x^2+i\epsilon}d(y)+\int_{-\pi/2}^{\pi/2} \frac{-i}{R^2e^{i2\phi}-x^2+i\epsilon}d(Re^{i\phi})\\
=\int_{\infty}^{-\infty}\frac{-i}{y^2-x^2+i\epsilon}dy+0\\
=\int_{-\infty}^{\infty}\frac{i}{y^2-x^2+i\epsilon}dy
$
Now lets use the residue theorem
$\oint_C \frac{-i}{z^2-x^2+i\epsilon}dz\\
=2\pi i \sum_{z^*\in C}\operatorname{Res}(\frac{-i}{z^2-x^2+i\epsilon},z^*)\\
=2\pi i\operatorname{Res}(\frac{-i}{z^2-x^2+i\epsilon},\sqrt{x^2-i\epsilon})\\
=2\pi i\lim(z-(\sqrt{x^2-i\epsilon}))\frac{-i}{z^2-x^2+i\epsilon}\\
=2\pi i\lim(z-(\sqrt{x^2-i\epsilon}))\frac{.i}{(z-\sqrt{x^2-i\epsilon})(z+\sqrt{x^2-i\epsilon})}\\
=2\pi i\lim_{z\to\sqrt{x^2-i\epsilon}}\frac{-i}{z+\sqrt{x^2-i\epsilon}}\\
=\frac{\pi}{\sqrt{x^2-i\epsilon}}\\
$
Finally lets caculate the entire integral
$\lim_{\epsilon\to 0^+}\int_{\Lambda/s}^\Lambda\int_{-\infty}^{\infty}\frac{i}{y^2-x^2+i\epsilon}dydx\\
=\lim_{\epsilon\to 0^+}\int_{\Lambda/s}^\Lambda\oint_C \frac{-i}{z^2-x^2+i\epsilon}dzdx\\
=\lim_{\epsilon\to 0^+}\int_{\Lambda/s}^\Lambda\frac{\pi}{\sqrt{x^2-i\epsilon}}dx\\
=\pi\lim_{\epsilon\to 0^+}\ln(\sqrt{x^2-i\epsilon}+x)|_{\Lambda/s}^\Lambda\\
=\pi\ln(2x)|_{\Lambda/s}^\Lambda\\
=\pi\ln(s)\\
$
$\therefore\lim_{\epsilon\to 0^+}\int_{\Lambda/s}^\Lambda \int_{-\infty}^\infty \frac{i}{y^2-x^2+i\epsilon} dydx=\pi\ln(s)$
For the last integral I just asked wolfram
Also I think that using $1/(y^2-x^2 +i \varepsilon)=PV (1/(y^2-x^2))-i \pi \delta(y^2-x^2)$ as mention in the comment it should be easier but I'm not sure how to apply it.
