Integral of $\int_{-1}^{1}x^{2n}e^{i k x}$

Question

Im trying to integrate \begin{equation} \int_{-1}^{1}x^{2n}\,\mathrm{e}^{\mathrm{i}kx}\,\mathrm{d}x = 2\left(\, -1\,\right)^{n}\, \frac{\mathrm{d}^{2n}}{\mathrm{d}k^{2n}}\left[\,{\frac{\sin\left(\, k\,\right)}{k}}\,\right] \end{equation}

I have the solution above. Now, I want to try contour integration to see if I get the same result.

How do I use contour integration?

Thanks in advance.

Answer 1

I hope I have understood your question correct.

I use the unit circle $\,|z|=1\,$ as contour.

For $z(t):=e^{it}$ and $\gamma=[-1,+1] $ we can write $$\int\limits_\gamma z^{k-1}(\ln z)^{2n}dz=\int\limits_{-1}^1 z(t)^{k-1}(\ln z(t))^{2n}z'(t)dt= \int\limits_{-1}^1 \frac{d^{2n}} {dk^{2n}}z(t)^{k-1}z'(t)dt$$ $$=\frac{d^{2n}} {dk^{2n}}\int\limits_{-1}^1 z(t)^{k-1}z'(t)dt=\frac{d^{2n}}{dk^{2n}}\frac{z(t)^k}{k}|_{-1}^{+1}=\frac{d^{2n}}{dk^{2n}}\frac{e^{ik}-e^{-ik}}{k}=i2\frac{d^{2n}}{dk^{2n}}\frac{\sin k}{k}$$ multiplicated with $-i(-1)^n$ it results in your equation.

Is this the answer to your question ?