What is the integral from 0 to 1 of e^{-(1+j^2pin)t} ?

The integral from 0 to 1 of e^{-(1+j^2pin)t} is (e^{-1-pij^2n}-1)/(-1-pij^2n)

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integral from 0 to 1 of e^{-(1+j^2pin)t}

\int_{0}^{1} e^{- (1 + j^{2} πn) t} d t

\frac{e ^{- 1 - π j^{2} n} - 1}{- 1 - π j ^{2} n}

Solution steps

\int_{0}^{1} e^{- (1 + j^{2} πn) t} d t

Apply u-substitution

: \int_{0}^{- 1 - π j^{2} n} \frac{e ^{u}}{- 1 - π j ^{2} n} d u

= \int_{0}^{- 1 - π j^{2} n} \frac{e ^{u}}{- 1 - π j ^{2} n} d u

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{- 1 - π j^{2} n} - 1

= \frac{e ^{- 1 - π j^{2} n} - 1}{- 1 - π j ^{2} n}

\frac{d}{d x} (sin (3 x) cos (3 x))

\int 2 cos (2 x) sin (2 x) d x

\int \frac{1}{( x ^{2} + 3 x + 5 ) ^{3}} d x

\int \frac{2}{3 x - 7} d x

x \to - 6 + lim (\frac{2 x}{6 + x})

What is the integral from 0 to 1 of e^{-(1+j^2pin)t} ?
The integral from 0 to 1 of e^{-(1+j^2pin)t} is (e^{-1-pij^2n}-1)/(-1-pij^2n)