What is the integral from 0 to infinity of e^{-xy}sin(x) ?

The integral from 0 to infinity of e^{-xy}sin(x) is 1/(y^2+1)

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integral from 0 to infinity of e^{-xy}sin(x)

\int_{0}^{\infty} e^{- x y} sin (x) d x

\frac{1}{y ^{2} + 1}

Solution steps

\int_{0}^{\infty} e^{- x y} sin (x) d x

Compute the indefinite integral:

\int e^{- x y} sin (x) d x = - \frac{e ^{- y x} cos ( x )}{1 + y ^{2}} - \frac{y e ^{- y x} sin ( x )}{1 + y ^{2}} + C

Compute the boundaries:

\int_{0}^{\infty} e^{- x y} sin (x) d x = 0 - (- \frac{1}{1 + y ^{2}})

= 0 - (- \frac{1}{1 + y ^{2}})

Simplify

= \frac{1}{y ^{2} + 1}

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

\int_{0}^{x} ln (1 + x) d x

\int_{0}^{0.3} x ln (1 + x^{3}) d x

\int_{1}^{2} (3 x^{3} + \frac{6}{x ^{2}}) d x

\int_{- 2}^{2} x^{5} d x

What is the integral from 0 to infinity of e^{-xy}sin(x) ?
The integral from 0 to infinity of e^{-xy}sin(x) is 1/(y^2+1)