Question Number 129576 by greg_ed last updated on 16/Jan/21
![please, how to show that f : [0 , a] × R_+ → R (x, y) e^(−xy) sin x is integrable ???](https://www.tinkutara.com/question/Q129576.png)
$$\boldsymbol{\mathrm{please}},\:\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{f}}\::\:\left[\mathrm{0}\:,\:\boldsymbol{{a}}\right]\:×\:\mathbb{R}_{+} \:\rightarrow\:\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{{x}},\:\boldsymbol{{y}}\right)\:\:\:\:\:\:\: \:\:\boldsymbol{{e}}^{−\boldsymbol{{xy}}} \:\boldsymbol{{sin}}\:\boldsymbol{{x}}\: \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{integrable}}\:???\: \\ $$
Answered by mathmax by abdo last updated on 18/Jan/21
![D=[0,a]×R^+ and ∫∫_D f(x,y)dxdy =∫_0 ^∞ (∫_0 ^a e^(−xy) sinxdx)dy =∫_0 ^∞ A(y)dy A(y)=∫_0 ^a e^(−yx) sinxdx =Im(∫_0 ^a e^(−yx+ix) dx) ∫_0 ^a e^((−y+i)x) dx =[(1/(−y+i)) e^((−y+i)x) ]_0 ^a =−(1/(y−i)){e^((−y+i)a) −1} =−((y+i)/(y^2 +1))(e^(−y) (cosa+isina)−1) =−(1/(y^2 +1))(y+i)(e^(−y) cosa+ie^(−y) sina−1) =−(1/(y^2 +1))(ye^(−y) cosa+iye^(−y) sina−y+ie^(−y) cosa−e^(−y) sina−i) ⇒ Im(...)=−(1/(y^2 +1))(ye^(−y) sina+e^(−y) cosa−1)=A(y) ⇒ ∫∫_D f(x,y)dxdy =−sina∫_0 ^∞ ((ye^(−y) )/(y^2 +1))dy −cosa∫_0 ^∞ (e^(−y) /(y^2 +1))dy+∫_0 ^∞ (dy/(1+y^2 )) ∫_0 ^∞ (dy/(1+y^2 ))=(π/2) ∫_0 ^∞ (e^(−y) /(1+y^2 ))dy ≤∫_0 ^∞ e^(−y) dy<+∞ ∃m>0 /∫_0 ^∞ (y/(y^2 +1))e^(−y) dy ≤m∫_0 ^∞ e^(−y) dy<+∞ ⇒f is integrable on D](https://www.tinkutara.com/question/Q129778.png)
$$\mathrm{D}=\left[\mathrm{0},\mathrm{a}\right]×\mathrm{R}^{+} \:\:\mathrm{and}\:\int\int_{\mathrm{D}} \mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\mathrm{dxdy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\mathrm{a}} \:\mathrm{e}^{−\mathrm{xy}} \mathrm{sinxdx}\right)\mathrm{dy}\:\:=\int_{\mathrm{0}} ^{\infty} \mathrm{A}\left(\mathrm{y}\right)\mathrm{dy} \\ $$$$\mathrm{A}\left(\mathrm{y}\right)=\int_{\mathrm{0}} ^{\mathrm{a}} \:\mathrm{e}^{−\mathrm{yx}} \:\mathrm{sinxdx}\:=\mathrm{Im}\left(\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{e}^{−\mathrm{yx}+\mathrm{ix}} \mathrm{dx}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{a}} \:\mathrm{e}^{\left(−\mathrm{y}+\mathrm{i}\right)\mathrm{x}} \mathrm{dx}\:=\left[\frac{\mathrm{1}}{−\mathrm{y}+\mathrm{i}}\:\mathrm{e}^{\left(−\mathrm{y}+\mathrm{i}\right)\mathrm{x}} \right]_{\mathrm{0}} ^{\mathrm{a}} \:=−\frac{\mathrm{1}}{\mathrm{y}−\mathrm{i}}\left\{\mathrm{e}^{\left(−\mathrm{y}+\mathrm{i}\right)\mathrm{a}} −\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{y}+\mathrm{i}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\left(\mathrm{e}^{−\mathrm{y}} \left(\mathrm{cosa}+\mathrm{isina}\right)−\mathrm{1}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\left(\mathrm{y}+\mathrm{i}\right)\left(\mathrm{e}^{−\mathrm{y}} \mathrm{cosa}+\mathrm{ie}^{−\mathrm{y}} \mathrm{sina}−\mathrm{1}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\left(\mathrm{ye}^{−\mathrm{y}} \:\mathrm{cosa}+\mathrm{iye}^{−\mathrm{y}} \mathrm{sina}−\mathrm{y}+\mathrm{ie}^{−\mathrm{y}} \mathrm{cosa}−\mathrm{e}^{−\mathrm{y}} \mathrm{sina}−\mathrm{i}\right)\:\Rightarrow \\ $$$$\mathrm{Im}\left(…\right)=−\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\left(\mathrm{ye}^{−\mathrm{y}} \mathrm{sina}+\mathrm{e}^{−\mathrm{y}} \mathrm{cosa}−\mathrm{1}\right)=\mathrm{A}\left(\mathrm{y}\right)\:\Rightarrow \\ $$$$\int\int_{\mathrm{D}} \mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\mathrm{dxdy}\:=−\mathrm{sina}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{ye}^{−\mathrm{y}} }{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dy}\:−\mathrm{cosa}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{y}} }{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dy}+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{y}} }{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\mathrm{dy}\:\leqslant\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{y}} \:\mathrm{dy}<+\infty \\ $$$$\exists\mathrm{m}>\mathrm{0}\:/\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{y}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{e}^{−\mathrm{y}} \:\mathrm{dy}\:\leqslant\mathrm{m}\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{y}} \:\mathrm{dy}<+\infty\:\Rightarrow\mathrm{f}\:\mathrm{is}\:\mathrm{integrable}\:\mathrm{on}\:\mathrm{D} \\ $$
Commented by greg_ed last updated on 20/Jan/21
$$\boldsymbol{\mathrm{thank}}\:\boldsymbol{\mathrm{u}}\:\boldsymbol{\mathrm{very}}\:\boldsymbol{\mathrm{much}},\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{mathmax}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{abdo}}\:! \\ $$