Upper bound for complex integral

Question

I am interested in obtaining a good upper bound for the absolute value of the following integral $$ \left| \int_{0}^{\pi/3} e^{-itn} \left( 1-e^{it} \right)^{k} dt \right|, $$ when $n>k>0$ are positive integers.

I can obtain an upper bound, $(2\pi)/((3(k+2))$, as follows:

• bringing the absolute value inside the integral,

• evaluating $\left| 1-e^{it} \right|$ in the range of $t$ as $2-2\cos(t)$,

• bounding $2-\cos(t)$ from above by $(3/\pi)t$,

• evaluating the resulting integral.

However, when I calculate the integral directly for various values of $n$ and $k$, I find that the resulting upper bound is larger than the value of the absolute value of the integral itself.

So I am wondering if people have ideas for how to get an improved upper bound.

Once I move the absolute value inside the integral, I lose a lot in my upper bound (and any dependence on $n$), so some other approach is needed.

Do people here have any ideas, references,…?

Answer 1

We want to upper-bound $|I_{n,k}|$, where $$I_{n,k}:=\int_0^{\pi/3} e^{-itn} (1-e^{it})^k\,dt.$$ Integrating by parts, we have $$I_{n,k}=\frac{e^{-itn}}{-in}(1-e^{it})^k\Big|_0^{\pi/3} -\int_0^{\pi/3} \frac{e^{-itn}}{-in}\,k(1-e^{it})^{k-1}(-ie^{it})\,dt.$$ So, $$n|I_{n,k}|\le|1-e^{i\pi/3}|^k+ \int_0^{\pi/3}k|1-e^{it}|^{k-1}\,dt.$$ Note that for $t\in(0,\pi/3)$ we have $|1-e^{it}|=2\sin\frac t2$, and hence $|1-e^{i\pi/3}|=1$ and $$\int_0^{\pi/3}k|1-e^{it}|^{k-1}\,dt =2^k\int_0^{1/2}k s^{k-1}\,\frac{ds}{\sqrt{1-s^2}} \\ \le2^k\int_0^{1/2}k s^{k-1}\,\frac{ds}{\sqrt{1-(1/2)^2}} =\frac1{\sqrt{1-(1/2)^2}}=\frac2{\sqrt3}.$$ So, $$|I_{n,k}|\le B_n:=\frac cn,$$ where $c:=1+\frac2{\sqrt3}.$

---

To illustrate the accuracy of the upper bound $B_n$ on $|I_{n,k}|$, here are the graphs $\{(n,k,r_{n,k})\colon n=2,\dots,30,\ k=1,\dots,n\}$ and $\{(n,r_{n,n-1})\colon n=2,\dots,30\}$ for $r_{n,k}:=\dfrac{|I_{n,k}|}{B_n}$, which suggest that $r_{n,k}$ is not less than something like $0.3$ for natural $n>k$.

Answer 2

Too long for a comment more than ten years later.

Starting from @Iosif Pinelis's nice answer $$f(k)=2^k\int_0^{\frac12}k\, s^{k-1}\,\frac{ds}{\sqrt{1-s^2}}=\, _2F_1\left(\frac{1}{2},\frac{k}{2};\frac{k+2}{2};\frac{1}{4}\right)$$

If we consider the function $$g(k)=k\, f(k)$$ for large values of $k$, its first derivative is $\frac{2}{\sqrt{3}}$ and all higher derivatives are $0$.

Asymptotically, $$g(k)=\frac{2}{\sqrt{3}}\left(k+\frac 13\right)$$