What is the integral from-1 to 1 of 2e^{iαx} ?

The integral from-1 to 1 of 2e^{iαx} is (4sin(α))/α

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integral from-1 to 1 of 2e^{iαx}

\int_{- 1}^{1} 2 e^{i αx} d x

\frac{4 sin ( α )}{α}

Solution steps

\int_{- 1}^{1} 2 e^{i αx} d x

Take the constant out:

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

= 2 \cdot \int_{- 1}^{1} e^{α i x} d x

Apply u-substitution

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

= 2 \cdot \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

= 2 (- \frac{i}{α} \cdot \int_{- i α}^{i α} e^{u} d u)

Use the common integral:

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

= 2 (- \frac{i}{α} [e^{u}]_{- i α}^{i α})

Simplify

= - 2 \cdot \frac{i}{α} [e^{u}]_{- i α}^{i α}

Compute the boundaries:

2 i sin (α)

= - 2 \cdot \frac{i}{α} \cdot 2 i sin (α)

Simplify

= \frac{4 sin ( α )}{α}

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What is the integral from-1 to 1 of 2e^{iαx} ?
The integral from-1 to 1 of 2e^{iαx} is (4sin(α))/α