# Proof of $\Re\int_0^{\infty}\exp(-ik-k/\sqrt{1-4ik})dk=\Im\int_0^{1/4}\exp(-k+ik/\sqrt{1-4k})dk$

As mentioned in the title, I want to show that $$\Re\int_0^{\infty}\exp\left(-ik-\frac{k}{\sqrt{1-4ik}}\right)dk=\Im\int_0^{1/4}\exp\left(-k+\frac{ik}{\sqrt{1-4k}}\right)dk.$$ I try to proof this equation by firstly closed the contour in the lower-half plane with a $$1/4$$ circle with radius $$R\to \infty$$, and the integral contour from $$-i\infty$$ to $$0$$ that avoids the singlarity at $$1/(4i)$$. According to the Jordan’s lemma, the integral contribution from the $$1/4$$ circle equals to $$0$$. Now I have to deal with a essential singlarity and have no any idea to proceed.

• @CarloBeenakker Idk how you perform these integrals. I seed these integrals to MMA and get 0.051997 for lhs and 0.051999 for rhs, with workingPrecision 200. Apr 14, 2023 at 11:33
• @CarloBeenakker I evaluate the lhs using MMA and get $-0.00395$ which matches your answer, but without setting the workprecision. Once I increase the WorkingPrecision I get $0.052$ that matches the rhs. Apr 14, 2023 at 11:42
• I think it's the high oscillatory term for large $k$ in the lhs, so it's necessary to increase the workprecision to decrease numerical errors. Apr 14, 2023 at 11:47
• clear enough, thanks. Apr 14, 2023 at 11:50
• I believe I know how to do it, but first I have to know what you mean by $\sqrt{1-4ik}$. Usually $\sqrt x$ is defined for $x\geq 0$ and it is equal to the number $y\geq 0$ such that $x=y^2$. If you want to extend the definition to complex numbers you have to specify which of the two branches of the radical you consider. If I am to guess, then perhaps you choose $\sqrt{1-4ik}$ to be the complex number $z$ with $1-4ik=z^2$ and $\Re z\geq 0$. Is that right? Apr 14, 2023 at 16:21

## 1 Answer

So your $$\sqrt z$$ is the "usual" branch of the square root, defined on $$\mathbb C\setminus (-\infty,0]$$. Here a number writes as $$z=r\exp(i\theta)$$, with $$r>0$$ and $$\theta\in (-\pi,\pi )$$, and $$\sqrt z:=\sqrt r\exp(i\theta/2)$$. You can extend the definition on the whole $$\mathbb C$$, but then you'll have discontinuity at negative numbers.

First, you don't have an essential singularity at $$1/4$$. Your function cannot be defined in the whole neighborhood of $$1/4$$, because $$\sqrt z$$ cannot be defined in the neighborhood of $$0$$. If you go clockwise around $$1/4$$ on a circle of radius $$\epsilon$$, starting and ending with $$1/4-\epsilon$$ then $$z=1/4+\epsilon\exp(i\theta)$$, with $$\theta\in (-\pi,\pi)$$, then your $$\sqrt{1-4z}$$ writes as $$\sqrt{-4\epsilon\exp(i\theta )}=\sqrt{4\epsilon\exp(i(\theta+\pi))}$$. When we make the change of variable $$\theta\to\theta+\pi$$, the circle parametrizes as $$z=1/4-\epsilon\exp(i\theta)$$ with $$\theta\in (0,2\pi)$$. So we need a new branch of the square root, defined on $$\mathbb C\setminus [0,\infty)$$. Here $$z=r\exp (i\theta)$$, with $$r>0$$ and $$\theta\in(0,2\pi)$$, and again we define $$\sqrt z:=\sqrt r\exp(i\theta/2)$$. To prevent confusions, we denote this new branch of the square root by $$\phi(z)$$.

How $$\phi (z)$$ differs from $$\sqrt z$$ on $$\mathbb C\setminus\mathbb R$$, where both functions are defined? If $$\Im z>0$$ then $$z=r\exp (i\theta)$$, with $$r>0$$ and $$\theta\in(0,\pi)$$. Since $$(0,\pi)$$ is included in both $$(-\pi,\pi)$$ and $$(0,2\pi)$$, we have $$\sqrt z=\phi (z)=\sqrt r\exp(i\theta/2)$$. If $$\Im z<0$$ then $$z=r\exp (i\theta)$$, with $$r>0$$ and $$\theta\in(-\pi,0)$$. We have $$\theta\in (-\pi,\pi )$$ so $$\sqrt z=\sqrt r\exp(i\theta/2)$$, but for $$\phi (z)$$ we write $$z=r\exp (i(\theta+2\pi))$$. Then $$\theta+2\pi\in(\pi,2\pi)\subset(0,2\pi)$$, so $$\phi(z)=\sqrt r\exp(i(\theta+2\pi)/2)=-\sqrt r\exp(i\theta/2)=-\sqrt z$$.

For short, $$\phi(z)=\sqrt z$$ if $$\Im z>0$$ and $$\phi(z)=-\sqrt z$$ if $$\Im z<0$$.

We now consider the function $$f(z)=\exp(-z-z/\phi(1-4z))$$. Then $$\phi(1-4z)$$ is defined for $$1-4z\notin[0,\infty)$$, i.e. $$\phi(1-4z)$$ is defined on $$\mathbb C\setminus(-\infty,1/4]$$ and so is $$f(z)$$.

If $$\Im z>0$$ then $$\Im(1-4z)<0$$ so $$\phi(1-4z)=-\sqrt{1-4z}$$ and $$f(z)=\exp(-z+z/\sqrt{1-4z})$$. If $$\Im z<0$$, then we get $$f(z)=\exp(-z-z/\sqrt{1-4z})$$.

Let $$\delta\ll\epsilon\ll 1\ll T$$. E consider $$\int f(z)dz$$ on the closed contour made of $$[i\delta,1/4-\epsilon+i\delta]$$, the arch of circle around $$1/4$$ of radius $$\sqrt{\epsilon^2+\delta^2}$$ around $$1/4$$ uniting $$1/4-\epsilon+i\delta$$ and $$1/4-\epsilon-i\delta$$, $$[1/4-\epsilon-i\delta,-i\delta]$$, $$[-i\delta,-iT]$$, the arch of circle of radius $$\sqrt{T^2+1/16}$$ around $$1/4$$ uniting $$-iT$$ and $$iT$$ and $$[iT,i\delta T]$$. This integral is $$0$$.

When we make $$\delta\to 0$$, the integral on $$[i\delta,1/4-\epsilon+i\delta]$$ becomes $$I_1=\int_0^{1/4-\epsilon}\exp(-k+ik/\sqrt{1-4k})dk$$. The integral on $$[1/4-\epsilon-i\delta,-i\delta]$$ becomes $$I_2=-\int_0^{1/4-\epsilon} \exp(-k+ik/\sqrt{1-4k})dk$$. The integral on $$[-i\delta,-iT]$$ becomes $$I_3=\int_0^Tf(-ik)d(-ik)=-i\int_0^T\exp(ik-k/\sqrt{1+4ik})dk$$. And the integral on $$[iT,i\delta]$$ becomes $$I_4=-\int_0^Tf(ik)d(ik)=-i\int_0^T\exp(-ik-k/\sqrt{1-4ik})dk$$. (Note that I used two different formulas for $$f(z)$$, one for $$\Im z>0$$, one for $$\Im z<0$$.)

We have $$I_2=-\bar I_1$$ so $$I_1+I_2=2i\Im I_1=2i\Im\int_0^{1/4-\epsilon}\exp(-k+ik/\sqrt{1-4k})dk$$. And if we put $$J=\int_0^T\exp(-ik-k/\sqrt{1-4ik})dk$$, then $$I_3=-i\bar J$$ and $$I_4=-iJ$$, so $$I_3+I_4=-i(J+\bar J)=-2i\Re J=-2i\Re\int_0^T\exp(-ik-k/\sqrt{1-4ik})dk$$.

In conclusion, we get $$0=2i\Im\int_0^{1/4-\epsilon}\exp(-k+ik/\sqrt{1-4k})dk-2i\Re\int_0^T\exp(-ik-k/\sqrt{1-4ik})dk+$$ the sum of two circular integrals. You have to prove that when $$\epsilon\to 0$$ and $$T\to\infty$$ the two circular integral go to zero. Then when you take the limits you get: $$0=2i\Im\int_0^{1/4}\exp(-k+ik/\sqrt{1-4k})dk-2i\Re\int_0^\infty\exp(-ik-k/\sqrt{1-4ik})dk$$ and you are done.

BTW when $$\delta\to 0$$ the small arch around $$1/4$$ of radius $$\sqrt{\epsilon^2+\delta^2}$$ becomes a full circle of radius $$\epsilon$$ around $$1/4$$, starting with $$1/4-\epsilon$$, in the negative sense.