What is the integral from 0 to infinity of sin(4x) ?

The integral from 0 to infinity of sin(4x) is diverges

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integral from 0 to infinity of sin(4x)

\int_{0}^{\infty} sin (4 x) d x

diverges

Solution steps

\int_{0}^{\infty} sin (4 x) d x

Compute the indefinite integral:

\int sin (4 x) d x = - \frac{1}{4} cos (4 x) + C

Compute the boundaries:

\int_{0}^{\infty} sin (4 x) d x =

diverges

= diverges

\int \frac{- 9}{x ^{2} + 16} d x

\frac{d ^{2}}{d x ^{2}} ((3 x^{2} + 6 x) sin (x))

\frac{d}{d x} (cos (2 x - 1) tan (1 - 2 x))

e^{x}, x e^{x^{2}}, [0, 1]

f (x) = \frac{1}{( 3 - x ) ^{2}}

What is the integral from 0 to infinity of sin(4x) ?
The integral from 0 to infinity of sin(4x) is diverges