What is the integral from-1 to 1 of e^{iαx} ?

The integral from-1 to 1 of e^{iαx} is (2sin(α))/α

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integral from-1 to 1 of e^{iαx}

\int_{- 1}^{1} e^{i αx} d x

\frac{2 sin ( α )}{α}

Solution steps

\int_{- 1}^{1} e^{i αx} d x

Apply u-substitution

: \int_{- i α}^{i α} - \frac{i e ^{u}}{α} d u

= \int_{- i α}^{i α} - \frac{i e ^{u}}{α} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{i}{α} \cdot \int_{- i α}^{i α} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{i}{α} [e^{u}]_{- i α}^{i α}

Compute the boundaries:

2 i sin (α)

= - \frac{i}{α} \cdot 2 i sin (α)

Simplify

= \frac{2 sin ( α )}{α}

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\int_{1}^{2} \frac{1 + t ^{2}}{t ^{4}} d t

What is the integral from-1 to 1 of e^{iαx} ?
The integral from-1 to 1 of e^{iαx} is (2sin(α))/α