Integral $\lim_{\epsilon\to 0^+} \int_{\Lambda/s}^\Lambda \int_{-\infty}^\infty \frac{i}{y^2-x^2+i\epsilon} dydx = \pi\log s$

Question

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?

Answer 1

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 $$\pi\int_{\Lambda/s}^{\Lambda}\frac{dx}{x} = \pi\log s\,. $$

Answer 2

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,

$$ \lim_{\epsilon \to 0^+} \int_{\Lambda/s}^{\Lambda} \int_{-\infty}^{\infty} \frac{i}{y^2 - x^2 + i\epsilon} \, \mathrm{d}y\mathrm{d}x = \lim_{\epsilon \to 0^+} \int_{\Lambda/s}^{\Lambda} \frac{\pi}{\sqrt{x^2 - i\epsilon}} \, \mathrm{d}x = \int_{\Lambda/s}^{\Lambda} \frac{\pi}{x} \, \mathrm{d}x = \pi \log s. $$

Answer 3

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.