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Question Number 129576 by greg_ed last updated on 16/Jan/21

please, how to show that

f : [0, a] \times R_{+} \to R

(x, y) e^{- x y} s i n x

is integrable ? ? ?

Answered by mathmax by abdo last updated on 18/Jan/21

D = [0, a] \times R^{+} and \int \int_{D} f (x, y) dxdy

= \int_{0}^{\infty} (\int_{0}^{a} e^{- xy} sinxdx) dy = \int_{0}^{\infty} A (y) dy

A (y) = \int_{0}^{a} e^{- yx} sinxdx = Im (\int_{0}^{a} e^{- yx + ix} dx)

\int_{0}^{a} e^{(- y + i) x} dx = {[\frac{1}{- y + i} e^{(- y + i) x}]}_{0}^{a} = - \frac{1}{y - i} {e^{(- y + i) a} - 1}

= - \frac{y + i}{y^{2} + 1} (e^{- y} (cosa + isina) - 1)

= - \frac{1}{y^{2} + 1} (y + i) (e^{- y} cosa + {ie}^{- y} sina - 1)

= - \frac{1}{y^{2} + 1} ({ye}^{- y} cosa + {iye}^{- y} sina - y + {ie}^{- y} cosa - e^{- y} sina - i) \Rightarrow

Im (\dots) = - \frac{1}{y^{2} + 1} ({ye}^{- y} sina + e^{- y} cosa - 1) = A (y) \Rightarrow

\int \int_{D} f (x, y) dxdy = - sina \int_{0}^{\infty} \frac{{ye}^{- y}}{y^{2} + 1} dy - cosa \int_{0}^{\infty} \frac{e^{- y}}{y^{2} + 1} dy + \int_{0}^{\infty} \frac{dy}{1 + y^{2}}

\int_{0}^{\infty} \frac{dy}{1 + y^{2}} = \frac{π}{2}

\int_{0}^{\infty} \frac{e^{- y}}{1 + y^{2}} dy ⩽ \int_{0}^{\infty} e^{- y} dy < + \infty

\exists m > 0 / \int_{0}^{\infty} \frac{y}{y^{2} + 1} e^{- y} dy ⩽ m \int_{0}^{\infty} e^{- y} dy < + \infty \Rightarrow f is integrable on D

Commented by greg_ed last updated on 20/Jan/21

thank u very much, sir mathmax by abdo!

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