What will this $\displaystyle \int_{-\frac{1}{4}}^{\frac{1}{4}}e^{2\pi ian }da$; $n\in\mathbb{Z}$ be equal to?
I know that $\displaystyle \int_{0}^{1}e^{2\pi ian }da \begin{cases} 1 & \text{ if } n=0 \\ 0 & \text{ if } n\neq 0 \end{cases}$. Or instead of $[0,1]$ we can take $[-\frac{1}{2},\frac{1}{2}]$.
Hints
For $n\neq 0$ we have
$$\int e^{2\pi i an}da=\frac{1}{2\pi i n}e^{2\pi i a n}+C$$
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$\sin\left(\frac{n\pi}{2}\right)=\begin{cases}0, \space \text{for n even}\\ (-1)^{\frac{n-1}{2}},\space\text{for n odd}\end{cases}$$
The case $n=0$ should be straightforward.
$$\int_{-\frac{1}{4}}^{\frac{1}{4}}e^{2\pi i an}da=\frac{1}{2\pi in}\left[e^{2\pi i an }\right]_{a=-\frac{1}{4}}^{a=\frac{1}{4}}=\frac{1}{2\pi i n}\left[e^{\frac{\pi in}{2}}-e^{-\frac{\pi i n}{2}}\right]=\frac{\sin\left(\frac{\pi n}{2} \right)}{\pi n}$$