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

Question

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.

Answer 1

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.