gamma function type complex integral estimation?

Question

Is there any way to compute or estimate the following integrals? Surely this type of thing must be known, but Wolfram and other online integrators are giving me either useless or zero information:

$$\int_{0}^{2\pi}x^n e^{\pm ix}dx$$ where $n\geq1$ is an integer.

Answer 1

Integration by parts gives $$ \begin{align} a_n &=\int_0^{2\pi}x^ne^{ix}\,\mathrm{d}x\tag{1}\\ &=-i(2\pi)^n+in\int_0^{2\pi}x^{n-1}e^{ix}\,\mathrm{d}x\tag{2}\\[6pt] &=-i(2\pi)^n+ina_{n-1}\tag{3}\\[6pt] &=i^nn!\sum_{k=1}^n\frac{(-i2\pi)^k}{ik!}\tag{4}\\ &=n!\sum_{k=1}^\infty\frac{(2\pi)^{n+k}(-i)^{k-1}}{(n+k)!}\tag{5} \end{align} $$ Explanation:
$(1)$: define $a_n$
$(2)$: integrate by parts
$(3)$: rewrite $(2)$ in terms of $(1)$
$(4)$: solve the recursion from $(3)$
$(5)$: use $\sum\limits_{k=1}^\infty\frac{(-i2\pi)^k}{k!}=e^{-2\pi i}-1=0$

The real and imaginary parts of $(5)$ give the asymptotic expansions $$ \int_0^{2\pi}x^n\cos(x)\,\mathrm{d}x=\frac{(2\pi)^{n+1}}{n+1}-\frac{(2\pi)^{n+3}}{(n+1)(n+2)(n+3)}+O\left(\frac{(2\pi)^n}{n^5}\right)\tag{6} $$ and $$ \int_0^{2\pi}x^n\sin(x)\,\mathrm{d}x=-\frac{(2\pi)^{n+2}}{(n+1)(n+2)}+\frac{(2\pi)^{n+4}}{(n+1)(n+2)(n+3)(n+4)}+O\left(\frac{(2\pi)^n}{n^6}\right)\tag{7} $$

Answer 2

Make the substitution $\mp ix=u$ and $\mp idx=du$.

$$\begin{align}\int_0^{2\pi}x^ne^{\pm ix}dx&=\pm i\int_0^{\mp2\pi i}(\mp iu)^ne^{-u}du\\ &=-(\mp i)^{n+1}\int_0^{\mp2\pi i}u^ne^{-u}du\\ &=-(\mp i)^{n+1}\gamma(n+1,\mp2\pi i) \end{align}$$

where $\gamma(n,x)$ is the incomplete gamma function.

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An example of an approximation:

$$\gamma(n,x)\approx x^n\left(\frac1n-\frac x{1!(n+1)}+\frac{x^2}{2!(n+2)}-\dots+\frac{(-x)^k}{k!(n+k)}\right)$$

for very large $k$.

Answer 3

We can write $$ \int_{\,x\, = \,0}^{\,2\pi } {x^{\,n\,} e^{\, \pm \,i\,x} \,dx} = \int_{\,x\, = \,0}^{\,2\pi } {x^{\,n\,} \left( {\cos x \pm \,i\sin x} \right)\,dx} $$ And we can start and compute the integral by parts $$ \begin{gathered} \int_{\,x\, = \,0}^{\,2\pi } {x^{\,n\,} \cos x\,dx} = \int_{\,x\, = \,0}^{\,2\pi } {x^{\,n\,} d\left( {\sin x} \right)} = \left. {\left( {x^{\,n\,} \sin x} \right)} \right|_{\,x\, = \,0}^{\,2\pi } - n\int_{\,x\, = \,0}^{\,2\pi } {x^{\,n - 1\,} \sin x\,dx} = \hfill \\ = - n\int_{\,x\, = \,0}^{\,2\pi } {x^{\,n - 1\,} \sin x\,dx} \hfill \\ \int_{\,x\, = \,0}^{\,2\pi } {x^{\,n\,} \sin x\,dx} = - \int_{\,x\, = \,0}^{\,2\pi } {x^{\,n\,} d\left( {\cos x} \right)} = \left. {\left( { - x^{\,n\,} \cos x} \right)} \right|_{\,x\, = \,0}^{\,2\pi } + n\int_{\,x\, = \,0}^{\,2\pi } {x^{\,n - 1\,} \cos x\,dx} = \hfill \\ = \left[ {0 = n} \right] - 2^{\,n\,} \pi ^{\,n\,} + n\int_{\,x\, = \,0}^{\,2\pi } {x^{\,n - 1\,} \cos x\,dx} \hfill \\ \end{gathered} $$ which translates into a double FDE.

Now, if your scope is to calculate the integrals on a computer, then you can use this result iteratively.