Question Number 53967 by maxmathsup by imad last updated on 27/Jan/19
$$\left.\mathrm{1}\right){calculate}\:{A}_{{t}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} \:{sinxdx}\:\:{with}\:{x}>\mathrm{0} \\ $$ $$\left.\mathrm{2}\right)\:{by}\:{using}\:{Fubuni}\:{theorem}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{x}}{dx}\:. \\ $$
Commented bymaxmathsup by imad last updated on 28/Jan/19
![1) we have A_t =Im(∫_0 ^∞ e^(−xt) e^(ix) dx) =Im(∫_0 ^∞ e^((i−t)x) dx) ∫_0 ^∞ e^((i−t)x) dx =[(1/(i−x)) e^((i−t)x) ]_(x=0) ^(x=+∞) = −(1/(i−t)) =(1/(t−i)) =((t+i)/(t^2 +1)) ⇒ A_t = (1/(t^2 +1)) 2) we have ∫_0 ^∞ A_t dt =∫_0 ^∞ (dt/(t^2 +1)) =[arctant]_0 ^(+∞) =(π/2) and by fubini ∫_0 ^∞ A_t dt =∫_0 ^∞ (∫_0 ^∞ e^(−xt) sinxdx)dt =∫_0 ^∞ (∫_0 ^∞ e^(−xt) dt)sinxdx =∫_0 ^∞ ([−(1/x) e^(−xt) ]_(t=0) ^(t=+∞) )sinx dx =∫_0 ^∞ ((sinx)/x) dx ⇒ ∫_0 ^∞ ((sinx)/x) dx =(π/2) .](Q54065.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{t}} ={Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} \:{e}^{{ix}} {dx}\right)\:={Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{\left({i}−{t}\right){x}} {dx}\right) \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left({i}−{t}\right){x}} {dx}\:=\left[\frac{\mathrm{1}}{{i}−{x}}\:{e}^{\left({i}−{t}\right){x}} \right]_{{x}=\mathrm{0}} ^{{x}=+\infty} \:=\:−\frac{\mathrm{1}}{{i}−{t}}\:=\frac{\mathrm{1}}{{t}−{i}}\:=\frac{{t}+{i}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$ $${A}_{{t}} =\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$ $$\left.\mathrm{2}\right)\:\:{we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:{A}_{{t}} {dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\left[{arctant}\right]_{\mathrm{0}} ^{+\infty} \:=\frac{\pi}{\mathrm{2}}\:\:{and}\:{by}\:{fubini} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:{A}_{{t}} {dt}\:=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} \:{sinxdx}\right){dt}\:=\int_{\mathrm{0}} ^{\infty} \:\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} {dt}\right){sinxdx} \\ $$ $$=\int_{\mathrm{0}} ^{\infty} \:\left(\left[−\frac{\mathrm{1}}{{x}}\:{e}^{−{xt}} \right]_{{t}=\mathrm{0}} ^{{t}=+\infty} \right){sinx}\:{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sinx}}{{x}}\:{dx}\:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{x}}\:{dx}\:=\frac{\pi}{\mathrm{2}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19
$${B}_{{t}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} {cosxdx} \\ $$ $${A}_{{t}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} {sinxdx} \\ $$ $${B}_{{t}} +{iA}_{{t}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} \left({cosx}+{isinx}\right){dx} \\ $$ $${B}_{{t}} +{iA}_{{t}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} .{e}^{{ix}} {dx}=\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}+{ix}} {dx} \\ $$ $$=\int_{\mathrm{0}} ^{\infty} {e}^{{x}\left(−{t}+{i}\right)} {dx}=\mid\frac{{e}^{{x}\left(−{t}+{i}\right)} }{−{t}+{i}}\mid_{\mathrm{0}} ^{\infty} \\ $$ $$=\mid\frac{{e}^{−{x}\left({t}−{i}\right)} }{−{t}+{i}}\mid_{\mathrm{0}} ^{\infty} =\frac{{e}^{−\infty\left({t}−{i}\right)} −{e}^{\mathrm{0}} }{−{t}+{i}}=\frac{−\mathrm{1}}{−{t}+{i}}=\frac{\mathrm{1}}{{t}−{i}} \\ $$ $$=\frac{{t}+{i}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+{i}×\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$ $${so}\:{B}_{{t}} =\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:\:\:{A}_{{t}} =\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19
$${C}_{{t}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} \:\frac{{sinx}}{{x}}{dx} \\ $$ $$\frac{{dC}_{{t}} }{{dt}}=\int_{\mathrm{0}} ^{\infty} \frac{\partial}{\partial{t}}\left(\frac{{e}^{−{xt}} {sinx}}{{x}}\right){dx} \\ $$ $$\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{xt}} ×−{x}×{sinx}}{{x}}{dx} \\ $$ $$\:\:\:\:\:\:\:\:\:\:=−\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} {sinxdx}=−{A}_{{t}} \\ $$ $$\frac{{dC}_{{t}} }{{dt}}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$ $$−{dC}_{{t}} =\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$ $$−{C}_{{t}} ={tan}^{−\mathrm{1}} \left({t}\right)+{k} \\ $$ $${k}=−{C}_{{t}} −{tan}^{−\mathrm{1}} \left({t}\right) \\ $$ $${when}\:{t}\rightarrow\infty\:\:{C}_{{t}} \rightarrow\mathrm{0}\:\:{and}\:{tan}^{−\mathrm{1}} \left({t}\right)\rightarrow\frac{\pi}{\mathrm{2}} \\ $$ $${so}\:{k}=−\frac{\pi}{\mathrm{2}} \\ $$ $$−{C}_{{t}} ={tan}^{−\mathrm{1}} \left({t}\right)−\frac{\pi}{\mathrm{2}} \\ $$ $${we}\:{have}\:{to}\:{find}\:\:\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx} \\ $$ $${we}\:{know}\:{that} \\ $$ $$−\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} \frac{{sinx}}{{x}}{dx}={tan}^{−\mathrm{1}} \left({t}\right)−\frac{\pi}{\mathrm{2}} \\ $$ $$ \\ $$ $${now}\:{put}\:{t}=\mathrm{0}\:{botb}\:{side} \\ $$ $$−\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}={tan}^{−\mathrm{1}} \left(\mathrm{0}\right)−\frac{\pi}{\mathrm{2}} \\ $$ $${so}\:\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}=\frac{\pi}{\mathrm{2}}\:\:{proved} \\ $$ $$ \\ $$