How to evaluate this integral $\int_{-a}^{a}{e^{-\iota k x}}dx$?

Question

I am having trouble to evaluate this integral.

$$\int_{-a}^{a}{e^{-i k x}}dx=2\frac{\sin\left(a k\right)}{k}$$

Here $\iota$ stands for imaginary part. Any input will be appreciated.

Answer 1

By Eular's formulate, $e^{-ikx} = cos(kx) -i \sin kx$

$$\int_{-a}^{a}{e^{-\iota k x}}dx= \int_{-a}^{a}{cos(kx)}dx - i\int_{-a}^{a}{ \sin kx}dx $$

From this post's first answer: since $ sin x $ is a ood function and the interval $[-a,a]$ symmetric about origin. $$\int_{-a}^{a}{e^{-\iota k x}}dx= \int_{-a}^{a}{cos(kx)}dx = 2sin(ka)/k$$

I think the a in the text is a typo.

Answer 2

Use:

• $$re^{\varphi i}=r\cos(\varphi)+r\sin(\varphi)i$$
• $$\cos(-x)=\cos(x)$$
• $$\sin(-x)=-\sin(x)$$

$$\int_{-a}^{a}e^{-ikx}\space\text{d}x=\int_{-a}^{a}\left(\cos(-kx)+\sin(-kx)i\right)\space\text{d}x=\int_{-a}^{a}\cos(kx)\space\text{d}x-i\int_{-a}^{a}\sin(kx)\space\text{d}x$$

Now, for the integrals substitute $u=kx$ and $\text{d}u=k\space\text{d}x$:

$$\frac{1}{k}\int_{-ak}^{ak}\cos(u)\space\text{d}u-\frac{i}{k}\int_{-ak}^{ak}\sin(u)\space\text{d}u$$

Now, use:

• $$\int\cos(x)\space\text{d}x=\sin(x)+\text{C}$$
• $$\int\sin(x)\space\text{d}x=\text{C}-\cos(x)$$

So, we get:

$$\frac{1}{k}\int_{-ak}^{ak}\cos(u)\space\text{d}u-\frac{i}{k}\int_{-ak}^{ak}\sin(u)\space\text{d}u=\frac{1}{k}\left[\sin(u)\right]_{-ak}^{ak}-\frac{i}{k}\left[-\cos(u)\right]_{-ak}^{ak}=$$ $$\frac{1}{k}\left(\sin(ak)-\sin(-ak)\right)+\frac{i}{k}\left(\cos(ak)-\cos(-ak)\right)=$$ $$\frac{1}{k}\left(\sin(ak)+\sin(ak)\right)+\frac{i}{k}\left(\cos(ak)-\cos(ak)\right)=$$ $$\frac{1}{k}\left(2\sin(ak)\right)+\frac{i}{k}\left(0\right)=\frac{2\sin(ak)}{k}+0=\frac{2\sin(ak)}{k}$$

Answer 3

Using the Taylor series of $e^{z} $ we get $$I=\int_{-a}^{a}e^{ikx}dx=\sum_{m\geq0}\frac{i^{m}k^{m}}{m!}\int_{-a}^{a}x^{m}dx=a\sum_{m\geq0}\frac{i^{m}k^{m}a^{m}}{\left(m+1\right)!}-a\sum_{m\geq0}\left(-1\right)^{m+1}\frac{i^{m}k^{m}a^{m}}{\left(m+1\right)!} $$ and now we note that there is canellation if $m $ is odd. So $$I=2a\sum_{m\geq0}\frac{i^{2m}k^{2m}a^{2m}}{\left(2m+1\right)!}=2a\sum_{m\geq0}\frac{\left(-1\right)^{m}k^{2m}a^{2m}}{\left(2m+1\right)!}=2\frac{\sin\left(ka\right)}{k}$$ from the Taylor series of $\sin$.