What is the integral from-1 to 0 of te^{-j^2pift} ?

The integral from-1 to 0 of te^{-j^2pift} is (-1-pifj^2e^{pifj^2}+e^{pifj^2})/(pi^2f^2j^4)

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integral from-1 to 0 of te^{-j^2pift}

\int_{- 1}^{0} t e^{- j^{2} π f t} d t

\frac{- 1 - π f j ^{2} e ^{π f j^{2}} + e ^{π f j^{2}}}{π ^{2} f ^{2} j ^{4}}

Solution steps

\int_{- 1}^{0} t e^{- j^{2} π f t} d t

Apply u-substitution

: \int_{π f j^{2}}^{0} \frac{e ^{u} u}{π ^{2} f ^{2} j ^{4}} d u

= \int_{π f j^{2}}^{0} \frac{e ^{u} u}{π ^{2} f ^{2} j ^{4}} d u

Take the constant out:

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

= \frac{1}{π ^{2} f ^{2} j ^{4}} \cdot \int_{π f j^{2}}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{π f j^{2}}^{0}

= \frac{1}{π ^{2} f ^{2} j ^{4}} [e^{u} u - \int e^{u} d u]_{π f j^{2}}^{0}

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

= \frac{1}{π ^{2} f ^{2} j ^{4}} [e^{u} u - e^{u}]_{π f j^{2}}^{0}

Simplify

\frac{1}{π ^{2} f ^{2} j ^{4}} [e^{u} u - e^{u}]_{π f j^{2}}^{0} : \frac{[ e ^{u} u - e ^{u} ] _{π f j^{2}}^{0}}{π ^{2} f ^{2} j ^{4}}

= \frac{[ e ^{u} u - e ^{u} ] _{π f j^{2}}^{0}}{π ^{2} f ^{2} j ^{4}}

Compute the boundaries:

- 1 - π f j^{2} e^{π f j^{2}} + e^{π f j^{2}}

= \frac{- 1 - π f j ^{2} e ^{π f j^{2}} + e ^{π f j^{2}}}{π ^{2} f ^{2} j ^{4}}

\frac{d}{d x} (6 x^{2} + 6 x - 36)

(x^{\frac{1}{2}})^{^{'}}

x \to 0 lim (a^{x} - b^{x})

\frac{d}{d x} (2 x \cdot ln (x))

\frac{d}{d x} (4 arctan (x))

What is the integral from-1 to 0 of te^{-j^2pift} ?
The integral from-1 to 0 of te^{-j^2pift} is (-1-pifj^2e^{pifj^2}+e^{pifj^2})/(pi^2f^2j^4)