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Popular Calculus >

integral of te^{-i2pift}

Solution

\int t e^{- i 2 π f t} d t

Solution

\frac{1}{4 π ^{2} f ^{2}} (cos (2 π f t) + 2 π f t sin (2 π f t)) + i \frac{1}{4 π ^{2} f ^{2}} (- sin (2 π f t) + 2 π f t cos (2 π f t)) + C

Solution steps

\int t e^{- i 2 π f t} d t

Apply u-substitution

= \int - \frac{e ^{u} u}{4 π ^{2} f ^{2}} d u

Take the constant out:

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

= - \frac{1}{4 π ^{2} f ^{2}} \cdot \int e^{u} u d u

Apply Integration By Parts

= - \frac{1}{4 π ^{2} f ^{2}} (e^{u} u - \int e^{u} d u)

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

= - \frac{1}{4 π ^{2} f ^{2}} (e^{u} u - e^{u})

Substitute back

u = - 2 f i t π

= - \frac{1}{4 π ^{2} f ^{2}} (e^{- 2 f i t π} (- 2 f i t π) - e^{- 2 f i t π})

Simplify

- \frac{1}{4 π ^{2} f ^{2}} (e^{- 2 f i t π} (- 2 f i t π) - e^{- 2 f i t π}) : \frac{1}{4 π ^{2} f ^{2}} (cos (2 π f t) + 2 π f t sin (2 π f t)) + i \frac{1}{4 π ^{2} f ^{2}} (- sin (2 π f t) + 2 π f t cos (2 π f t))

= \frac{1}{4 π ^{2} f ^{2}} (cos (2 π f t) + 2 π f t sin (2 π f t)) + i \frac{1}{4 π ^{2} f ^{2}} (- sin (2 π f t) + 2 π f t cos (2 π f t))

Add a constant to the solution

= \frac{1}{4 π ^{2} f ^{2}} (cos (2 π f t) + 2 π f t sin (2 π f t)) + i \frac{1}{4 π ^{2} f ^{2}} (- sin (2 π f t) + 2 π f t cos (2 π f t)) + C

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