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

The integral from-1 to 1 of e^{-jwt} is -(e^{-wj}-e^{wj})/(wj)

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integral from-1 to 1 of e^{-jwt}

\int_{- 1}^{1} e^{- j wt} d t

- \frac{e ^{- w j} - e ^{w j}}{w j}

Solution steps

\int_{- 1}^{1} e^{- j wt} d t

Apply u-substitution

: \int_{w j}^{- w j} - \frac{e ^{u}}{w j} d u

= \int_{w j}^{- w j} - \frac{e ^{u}}{w j} d u

Take the constant out:

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

= - \frac{1}{w j} \cdot \int_{w j}^{- w j} e^{u} d u

Use the common integral:

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

= - \frac{1}{w j} [e^{u}]_{w j}^{- w j}

Simplify

- \frac{1}{w j} [e^{u}]_{w j}^{- w j} : - \frac{[ e ^{u} ] _{w j}^{- w j}}{w j}

= - \frac{[ e ^{u} ] _{w j}^{- w j}}{w j}

Compute the boundaries:

e^{- w j} - e^{w j}

= - \frac{e ^{- w j} - e ^{w j}}{w j}

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What is the integral from-1 to 1 of e^{-jwt} ?
The integral from-1 to 1 of e^{-jwt} is -(e^{-wj}-e^{wj})/(wj)