How to integrate $\int \frac{e^{-ikt}}{\sqrt{1+it}}\,dt$ numerically?

Question

How should one numerically integrate a complex exponential function like

$$\int \frac{e^{-ikt}}{\sqrt{1+it}} dt$$

where it is not obvious how to separate the real and imaginary parts of integrand?

Using Euler's formula:

$$ \sum \frac{\cos (kt) - i \sin (kt)}{\sqrt{1+it}} \Delta t$$

seems to require us, first of all, to be able to separate (eg. this integral) the real and imaginary parts cleanly before applying numerical integration?

Answer 1

Let $$\sqrt{1+it}=a+ib$$

where $a=a(t), b=b(t)$ are real functions.

Square it, we get $$1+it=a^2-b^2+2abi$$

Compare the real and imaginary part, we get

$$1=a^2-b^2,~~~t=2ab$$

Solve and we get

$$a=\sqrt{\frac{1+\sqrt{1+t^2}}{2}},~~b=\sqrt{\frac{-1+\sqrt{1+t^2}}{2}}$$

$$$$ Therefore,

$$\begin{align}\frac{1}{\sqrt{1+it}}&=\frac{1}{a+ib}=\frac{a-ib}{a^2+b^2}\\ \\ &=\frac{1}{\sqrt{1+t^2}}\sqrt{\frac{1+\sqrt{1+t^2}}{2}}-i\frac{1}{\sqrt{1+t^2}}\sqrt{\frac{-1+\sqrt{1+t^2}}{2}}\end{align}$$

Further,

$$\begin{align}\frac{e^{-ikt}}{\sqrt{1+it}}&=\left(\sqrt{\frac{1+\sqrt{1+t^2}}{2(1+t^2)}}-i\sqrt{\frac{-1+\sqrt{1+t^2}}{2(1+t^2)}}\right)(\cos kt-i\sin kt)\\ \\ &=R(t)+i\cdot I(t)\end{align}$$

where

$$\begin{align}R(t)&=\cos kt\cdot\sqrt{\frac{1+\sqrt{1+t^2}}{2(1+t^2)}}-\sin kt\cdot\sqrt{\frac{-1+\sqrt{1+t^2}}{2(1+t^2)}}\\ \\ I(t)&=-\cos kt\cdot\sqrt{\frac{-1+\sqrt{1+t^2}}{2(1+t^2)}}-\sin kt \cdot\sqrt{\frac{1+\sqrt{1+t^2}}{2(1+t^2)}}\end{align}$$

Hence, your integral becomes:

$$\int \frac{e^{-ikt}}{\sqrt{1+it}} dt=\int R(t) dt+i\int I(t) dt$$

Now you can do the numerical integration on these two real integrals.