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Calculate-0-sin-z-z-z-2-1-dz-method-1-using-Laplace-Transform-method-2-using-Contour-Integral-method-3-using-Feymann-s-parametirc-trick-HELP-

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Question Number 195421 by MathedUp last updated on 02/Aug/23
Calculate. ∫_0 ^( ∞) ((sin(z))/(z(z^2 +1)))dz method 1. using Laplace Transform. method 2. using Contour Integral. method 3. using Feymann′s parametirc trick HELP!!!!!!!
$$\mathrm{Calculate}.\:\:\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}\left({z}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{d}{z} \\ $$$$\mathrm{method}\:\mathrm{1}.\:\mathrm{using}\:\mathrm{Laplace}\:\mathrm{Transform}. \\ $$$$\mathrm{method}\:\mathrm{2}.\:\mathrm{using}\:\mathrm{Contour}\:\mathrm{Integral}. \\ $$$$\mathrm{method}\:\mathrm{3}.\:\mathrm{using}\:\mathrm{Feymann}'\mathrm{s}\:\mathrm{parametirc}\:\mathrm{trick} \\ $$$$\mathrm{HELP}!!!!!!! \\ $$
Answered by witcher3 last updated on 02/Aug/23
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Commented by witcher3 last updated on 02/Aug/23
Small ΩCircle radius ε Big Γ radius R L=[−R,−ε],L′=[ε,R] Γ=Re^(ia) ,a∈[0,π] c=εe^(ia) ,a∈[0,π] D=Ω∪Γ∪L∪L′ f(z)=(e^(iz) /(z(1+z^2 ))),hase Twos pols in D z^2 +1=0,z=i ∫_D f(z)dz=2iπRes(f,i,)=2iπ.(e^(−1) /(i(2i)))=(π/(ie)) ∫_(−R) ^(−ε) f(z)dz+∫_π ^0 f(εe^(ia) ).iεe^(ia) da+∫_ε ^R f(z)dz+∫_0 ^π f(Re^(ia) )iRe^(ia) da ∫_(−R) ^(−ε) f(z)dz+∫_ε ^R f(z)dz=∫_ε ^R (f(z)+f(−z))dz =2i∫_ε ^R ((sin(z))/(z(1+z^2 )))dz lim_(ε→0) ∫_π ^0 f(εe^(ia) )iεe^(ia) da=lim_(ε→0) i∫_π ^0 (e^(iεe^(ia) ) /((1+ε^2 e^(2ia) )))da integral Cv uniformaly =i∫_π ^0 lim_(ε→0) (e^(iεe^(ia) ) /(1+ε^2 e^(2ia) ))da=−iπ lim_(ε→0) ∫_Ω f(z)dz=−iπ ∫_0 ^π f(Re^(ia) )iRe^(ia) da=i∫_0 ^π (e^(iRe^(ia) ) /(1+R^2 e^(2ia) ))da =i∫_0 ^π ((e^(iRcos(a)) .e^(−Rsin(a)) )/(1+R^2 e^(2ia) )) p ∣((e^(iRcos(a)) .e^(−Rsin(a)) )/(1+R^2 e^(2ia) ))∣≤(e^(−Rsin(a)) /(R^2 −1))⇒ =lim_(R→∞) ∣i∫_0 ^π ((e^(iRcos(a)) .e^(−Rsin(a)) )/(1+R^2 e^(2ia) ))∣da≤∫_0 ^π lim_(R→∞) (e^(−Rsin(a)) /(R^2 −1))da =0 lim_(R→∞) ∫_Γ f(z)dz=0 lim_(ε→0,R→∞) ∫_D f(z)dz=2i∫_0 ^∞ ((sin(z))/(z(1+z^2 )))dz−iπ+0=(π/(ie)) ∫_0 ^∞ ((sin(x))/(x(1+x^2 )))dx=(π/2)−(π/(2e))=(π/2)(((e−1)/e))
$$\mathrm{Small}\:\Omega\mathrm{Circle}\:\mathrm{radius}\:\epsilon \\ $$$$\mathrm{Big}\:\:\Gamma\:\mathrm{radius}\:\mathrm{R} \\ $$$$\mathrm{L}=\left[−\mathrm{R},−\epsilon\right],\mathrm{L}'=\left[\epsilon,\mathrm{R}\right] \\ $$$$\Gamma=\mathrm{Re}^{\mathrm{ia}} ,\mathrm{a}\in\left[\mathrm{0},\pi\right] \\ $$$$\mathrm{c}=\epsilon\mathrm{e}^{\mathrm{ia}} ,\mathrm{a}\in\left[\mathrm{0},\pi\right] \\ $$$$\mathrm{D}=\Omega\cup\Gamma\cup\mathrm{L}\cup\mathrm{L}' \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)},\mathrm{hase}\:\mathrm{Twos}\:\mathrm{pols}\:\mathrm{in}\:\mathrm{D} \\ $$$$\mathrm{z}^{\mathrm{2}} +\mathrm{1}=\mathrm{0},\mathrm{z}=\mathrm{i} \\ $$$$\int_{\mathrm{D}} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\mathrm{Res}\left(\mathrm{f},\mathrm{i},\right)=\mathrm{2i}\pi.\frac{\mathrm{e}^{−\mathrm{1}} }{\mathrm{i}\left(\mathrm{2i}\right)}=\frac{\pi}{\mathrm{ie}} \\ $$$$\int_{−\mathrm{R}} ^{−\epsilon} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}+\int_{\pi} ^{\mathrm{0}} \mathrm{f}\left(\epsilon\mathrm{e}^{\mathrm{ia}} \right).\mathrm{i}\epsilon\mathrm{e}^{\mathrm{ia}} \mathrm{da}+\int_{\epsilon} ^{\mathrm{R}} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}+\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{Re}^{\mathrm{ia}} \right)\mathrm{iRe}^{\mathrm{ia}} \mathrm{da} \\ $$$$\int_{−\mathrm{R}} ^{−\epsilon} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}+\int_{\epsilon} ^{\mathrm{R}} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}=\int_{\epsilon} ^{\mathrm{R}} \left(\mathrm{f}\left(\mathrm{z}\right)+\mathrm{f}\left(−\mathrm{z}\right)\right)\mathrm{dz} \\ $$$$=\mathrm{2i}\int_{\epsilon} ^{\mathrm{R}} \frac{\mathrm{sin}\left(\mathrm{z}\right)}{\mathrm{z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}\mathrm{dz} \\ $$$$\underset{\epsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{\pi} ^{\mathrm{0}} \mathrm{f}\left(\epsilon\mathrm{e}^{\mathrm{ia}} \right)\mathrm{i}\epsilon\mathrm{e}^{\mathrm{ia}} \mathrm{da}=\underset{\epsilon\rightarrow\mathrm{0}} {\mathrm{lim}i}\int_{\pi} ^{\mathrm{0}} \frac{\mathrm{e}^{\mathrm{i}\epsilon\mathrm{e}^{\mathrm{ia}} } }{\left(\mathrm{1}+\epsilon^{\mathrm{2}} \mathrm{e}^{\mathrm{2ia}} \right)}\mathrm{da} \\ $$$$\mathrm{integral}\:\mathrm{Cv}\:\mathrm{uniformaly} \\ $$$$=\mathrm{i}\int_{\pi} ^{\mathrm{0}} \underset{\epsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{i}\epsilon\mathrm{e}^{\mathrm{ia}} } }{\mathrm{1}+\epsilon^{\mathrm{2}} \mathrm{e}^{\mathrm{2ia}} }\mathrm{da}=−\mathrm{i}\pi \\ $$$$\underset{\epsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{\Omega} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}=−\mathrm{i}\pi \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{Re}^{\mathrm{ia}} \right)\mathrm{iRe}^{\mathrm{ia}} \mathrm{da}=\mathrm{i}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{e}^{\mathrm{iRe}^{\mathrm{ia}} } }{\mathrm{1}+\mathrm{R}^{\mathrm{2}} \mathrm{e}^{\mathrm{2ia}} }\mathrm{da} \\ $$$$=\mathrm{i}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{e}^{\mathrm{iRcos}\left(\mathrm{a}\right)} .\mathrm{e}^{−\mathrm{Rsin}\left(\mathrm{a}\right)} }{\mathrm{1}+\mathrm{R}^{\mathrm{2}} \mathrm{e}^{\mathrm{2ia}} } \\ $$$$\mathrm{p} \\ $$$$\mid\frac{\mathrm{e}^{\mathrm{iRcos}\left(\mathrm{a}\right)} .\mathrm{e}^{−\mathrm{Rsin}\left(\mathrm{a}\right)} }{\mathrm{1}+\mathrm{R}^{\mathrm{2}} \mathrm{e}^{\mathrm{2ia}} }\mid\leqslant\frac{\mathrm{e}^{−\mathrm{Rsin}\left(\mathrm{a}\right)} }{\mathrm{R}^{\mathrm{2}} −\mathrm{1}}\Rightarrow \\ $$$$=\underset{\mathrm{R}\rightarrow\infty} {\mathrm{lim}}\mid\mathrm{i}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{e}^{\mathrm{iRcos}\left(\mathrm{a}\right)} .\mathrm{e}^{−\mathrm{Rsin}\left(\mathrm{a}\right)} }{\mathrm{1}+\mathrm{R}^{\mathrm{2}} \mathrm{e}^{\mathrm{2ia}} }\mid\mathrm{da}\leqslant\int_{\mathrm{0}} ^{\pi} \underset{\mathrm{R}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{e}^{−\mathrm{Rsin}\left(\mathrm{a}\right)} }{\mathrm{R}^{\mathrm{2}} −\mathrm{1}}\mathrm{da} \\ $$$$=\mathrm{0} \\ $$$$\underset{\mathrm{R}\rightarrow\infty} {\mathrm{lim}}\int_{\Gamma} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{0} \\ $$$$ \\ $$$$\underset{\epsilon\rightarrow\mathrm{0},\mathrm{R}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{D}} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{z}\right)}{\mathrm{z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}\mathrm{dz}−\mathrm{i}\pi+\mathrm{0}=\frac{\pi}{\mathrm{ie}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2e}}=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{e}−\mathrm{1}}{\mathrm{e}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by senestro last updated on 02/Aug/23
(π/2)(1+(1/e))
$$\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{{e}}\right) \\ $$
Answered by witcher3 last updated on 02/Aug/23
Laplace ∫_0 ^∞ ((sin(zt))/(z(1+z^2 )))dz=f(t) L(f(t))=∫_0 ^∞ f(t)e^(−ts) dt=∫_0 ^∞ ∫_0 ^∞ ((sin(zt))/(z(1+z^2 )))dze^(−ts) dt =∫_0 ^∞ (1/(z(1+z^2 )))∫_0 ^∞ sin(zt)e^(−ts) dtdz =∫_0 ^∞ (1/(z(1+z^2 )))L{sin(zt)}dz =∫_0 ^∞ (1/(z(1+z^2 ))).(z/(z^2 +s^2 ))dz=∫_0 ^∞ (dz/((1+z^2 )(z^2 +s^2 ))) =(1/2)∫_(−∞) ^∞ (dz/((1+z^2 )(z^2 +s^2 )))=iπ.(1/(2i(s^2 −1)))+iπ.(1/(2is(1−s^2 ))) =(π/(2s(1+s)))=(π/(2s))−(π/(2(1+s))) f(t)=L^− (L(f(t))=L^− ((π/2)((1/s)−(1/(s+1)))) =(π/2)L^− ((1/s))−(π/2)L^− ((1/(s+1)))=(π/2)−(π/2)e^(−t) f(t)=∫_0 ^∞ ((sin(xt))/(x(1+x^2 )))dx=(π/2)(1−e^(−t) ) ∫_0 ^∞ ((sin(x))/(x(1+x^2 )))dx=f(1)=(π/2)(1−(1/e))
$$\mathrm{Laplace} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{zt}\right)}{\mathrm{z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}\mathrm{dz}=\mathrm{f}\left(\mathrm{t}\right) \\ $$$$\mathcal{L}\left(\mathrm{f}\left(\mathrm{t}\right)\right)=\int_{\mathrm{0}} ^{\infty} \mathrm{f}\left(\mathrm{t}\right)\mathrm{e}^{−\mathrm{ts}} \mathrm{dt}=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{zt}\right)}{\mathrm{z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}\mathrm{dze}^{−\mathrm{ts}} \mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}\int_{\mathrm{0}} ^{\infty} \mathrm{sin}\left(\mathrm{zt}\right)\mathrm{e}^{−\mathrm{ts}} \mathrm{dtdz} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}\mathcal{L}\left\{\mathrm{sin}\left(\mathrm{zt}\right)\right\}\mathrm{dz} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}.\frac{\mathrm{z}}{\mathrm{z}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} }\mathrm{dz}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dz}}{\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{\mathrm{dz}}{\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} \right)}=\mathrm{i}\pi.\frac{\mathrm{1}}{\mathrm{2i}\left(\mathrm{s}^{\mathrm{2}} −\mathrm{1}\right)}+\mathrm{i}\pi.\frac{\mathrm{1}}{\mathrm{2is}\left(\mathrm{1}−\mathrm{s}^{\mathrm{2}} \right)} \\ $$$$=\frac{\pi}{\mathrm{2s}\left(\mathrm{1}+\mathrm{s}\right)}=\frac{\pi}{\mathrm{2s}}−\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\mathrm{s}\right)} \\ $$$$\mathrm{f}\left(\mathrm{t}\right)=\mathcal{L}^{−} \left(\mathcal{L}\left(\mathrm{f}\left(\mathrm{t}\right)\right)=\mathcal{L}^{−} \left(\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{s}}−\frac{\mathrm{1}}{\mathrm{s}+\mathrm{1}}\right)\right)\right. \\ $$$$=\frac{\pi}{\mathrm{2}}\mathcal{L}^{−} \left(\frac{\mathrm{1}}{\mathrm{s}}\right)−\frac{\pi}{\mathrm{2}}\mathcal{L}^{−} \left(\frac{\mathrm{1}}{\mathrm{s}+\mathrm{1}}\right)=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{t}} \\ $$$$\mathrm{f}\left(\mathrm{t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{xt}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{t}} \right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}=\mathrm{f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}}\right) \\ $$$$ \\ $$
Answered by witcher3 last updated on 02/Aug/23
methode (3) y(t)=∫_0 ^∞ ((sin(xt))/(x(1+x^2 )))dx,t>0 y(0)=0 y′(t)=∫_0 ^∞ ((cos(xt))/(1+x^2 ))dx,y′(0)=(π/2) y′′(t)=∫_0 ^∞ ((−x^2 sin(xt))/(x(1+x^2 )))dx Y′′(t)−Y(t)=∫_0 ^∞ ((−x^2 sin(xt))/(x(1+x^2 )))−((sin(xt))/(x(1+x^2 )))dx =−∫_0 ^∞ ((sin(xt))/x)dx=−∫_0 ^∞ ((sin(xt))/(xt))d(xt) =−∫_0 ^∞ ((sin(z))/z)dz=−(π/2) y′′(t)−y(t)=−(π/2) y(t)=ae^t +be^(−t) +(π/2) y(0)=0,y′(0)=(π/2)⇒ { ((a+b=−(π/2))),((a−b=(π/2))) :}⇒a=0,b=−(π/2) y(t)=(π/2)(1−(1/e^t )),y(1)=∫_0 ^∞ ((sin(x))/(x(1+x^2 )))=(π/2)(1−(1/e))
$$\mathrm{methode}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{y}\left(\mathrm{t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{xt}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx},\mathrm{t}>\mathrm{0} \\ $$$$\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{y}'\left(\mathrm{t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\left(\mathrm{xt}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx},\mathrm{y}'\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{y}''\left(\mathrm{t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{−\mathrm{x}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{xt}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$\mathrm{Y}''\left(\mathrm{t}\right)−\mathrm{Y}\left(\mathrm{t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{−\mathrm{x}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{xt}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}−\frac{\mathrm{sin}\left(\mathrm{xt}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{xt}\right)}{\mathrm{x}}\mathrm{dx}=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{xt}\right)}{\mathrm{xt}}\mathrm{d}\left(\mathrm{xt}\right) \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{z}\right)}{\mathrm{z}}\mathrm{dz}=−\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{y}''\left(\mathrm{t}\right)−\mathrm{y}\left(\mathrm{t}\right)=−\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{y}\left(\mathrm{t}\right)=\mathrm{ae}^{\mathrm{t}} +\mathrm{be}^{−\mathrm{t}} +\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0},\mathrm{y}'\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\Rightarrow \\ $$$$\begin{cases}{\mathrm{a}+\mathrm{b}=−\frac{\pi}{\mathrm{2}}}\\{\mathrm{a}−\mathrm{b}=\frac{\pi}{\mathrm{2}}}\end{cases}\Rightarrow\mathrm{a}=\mathrm{0},\mathrm{b}=−\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{y}\left(\mathrm{t}\right)=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{t}} }\right),\mathrm{y}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}}\right) \\ $$$$ \\ $$
Commented by MathedUp last updated on 03/Aug/23
i think you are god of math LOL
$${i}\:{think}\:{you}\:{are}\:{god}\:{of}\:{math}\:{LOL} \\ $$
Commented by witcher3 last updated on 03/Aug/23
no Just i just orck Hard and the secret is always worck hard bro do your best and stop social media! evrey one withe enough worck progress in evrey steps of life, in this forum they are many people Good in Maths &Physics God bless You Sir “please Dont use God for a person ”
$$\mathrm{no}\:\mathrm{Just}\:\mathrm{i}\:\mathrm{just}\:\mathrm{orck}\:\mathrm{Hard}\:\mathrm{and}\:\mathrm{the}\:\mathrm{secret}\:\mathrm{is}\:\mathrm{always}\: \\ $$$$\mathrm{worck}\:\mathrm{hard}\:\mathrm{bro}\:\mathrm{do}\:\mathrm{your}\:\mathrm{best}\:\mathrm{and}\:\mathrm{stop}\:\mathrm{social}\:\mathrm{media}! \\ $$$$\mathrm{evrey}\:\mathrm{one}\:\mathrm{withe}\:\mathrm{enough}\:\mathrm{worck}\:\mathrm{progress}\:\mathrm{in}\:\mathrm{evrey}\: \\ $$$$\mathrm{steps}\:\mathrm{of}\:\mathrm{life},\: \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{forum}\:\mathrm{they}\:\mathrm{are}\:\mathrm{many}\:\mathrm{people}\:\mathrm{Good}\: \\ $$$$\mathrm{in}\:\mathrm{Maths}\:\&\mathrm{Physics}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{You}\:\mathrm{Sir} \\ $$$$“\mathrm{please}\:\mathrm{Dont}\:\mathrm{use}\:\mathrm{God}\:\mathrm{for}\:\mathrm{a}\:\mathrm{person}\:'' \\ $$
Commented by MM42 last updated on 03/Aug/23
peace be upon you .you are a person with a personality.
$${peace}\:{be}\:{upon}\:{you}\:.{you}\:{are}\:{a}\: \\ $$$${person}\:{with}\:{a}\:{personality}. \\ $$
Commented by witcher3 last updated on 09/Aug/23
thank You God bless You
$$\mathrm{thank}\:\mathrm{You}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{You} \\ $$

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