What is the integral of e^{itx} ?

The integral of e^{itx} is 1/t sin(tx)-i 1/t cos(tx)+C

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integral of e^{itx}

\int e^{i t x} d x

\frac{1}{t} sin (t x) - i \frac{1}{t} cos (t x) + C

Solution steps

\int e^{i t x} d x

Apply u-substitution

: \int - \frac{i e ^{u}}{t} d u

= \int - \frac{i e ^{u}}{t} d u

Take the constant out:

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

= - \frac{i}{t} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{i}{t} e^{u}

Substitute back

u = t i x

= - \frac{i}{t} e^{t i x}

Simplify

- \frac{i}{t} e^{t i x} : \frac{1}{t} sin (t x) - i \frac{1}{t} cos (t x)

= \frac{1}{t} sin (t x) - i \frac{1}{t} cos (t x)

Add a constant to the solution

= \frac{1}{t} sin (t x) - i \frac{1}{t} cos (t x) + C

x \to 2 lim (\frac{5 x}{x - 3})

\frac{\partial}{\partial z} (e^{x z})

\int (ln (x))^{4} d x

f (x) = 2 + x^{2}

n = 0 \sum \infty \frac{n x ^{n}}{n + 1}

What is the integral of e^{itx} ?
The integral of e^{itx} is 1/t sin(tx)-i 1/t cos(tx)+C