What is the integral of te^{-jwt} ?

The integral of te^{-jwt} is 1/(w^2j^2)(-wje^{-wjt}t-e^{-wjt})+C

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integral of te^{-jwt}

\int t e^{- j wt} d t

\frac{1}{w ^{2} j ^{2}} (- w j e^{- w j t} t - e^{- w j t}) + C

Solution steps

\int t e^{- j wt} d t

Apply u-substitution

= \int \frac{ln ( u )}{w ^{2} j ^{2}} d u

Take the constant out:

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

= \frac{1}{w ^{2} j ^{2}} \cdot \int ln (u) d u

Apply Integration By Parts

= \frac{1}{w ^{2} j ^{2}} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{w ^{2} j ^{2}} (u ln (u) - u)

Substitute back

u = e^{- w j t}

= \frac{1}{w ^{2} j ^{2}} (e^{- w j t} ln (e^{- w j t}) - e^{- w j t})

e^{- w j t} ln (e^{- w j t}) = - w j e^{- w j t} t

= \frac{1}{w ^{2} j ^{2}} (- w j e^{- w j t} t - e^{- w j t})

Add a constant to the solution

= \frac{1}{w ^{2} j ^{2}} (- w j e^{- w j t} t - e^{- w j t}) + C

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What is the integral of te^{-jwt} ?
The integral of te^{-jwt} is 1/(w^2j^2)(-wje^{-wjt}t-e^{-wjt})+C