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Question Number 205775 by SANOGO last updated on 30/Mar/24
calcu/ limit/n→+oo ∫_0 ^(+oo) arctan((x/n))e^(−x) dx
$${calcu}/\:\:\:\:{limit}/{n}\rightarrow+{oo} \\ $$$$\:\:\int_{\mathrm{0}} ^{+{oo}} {arctan}\left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx} \\ $$
Answered by MathedUp last updated on 30/Mar/24
let′s consider F(s)=∫_0 ^(∞ ) tan^(−1) (t)e^(−st) dt ∙∙∙∙A L_s =∫_0 ^∞ e^(−st) ∙ (Laplace Transform) F(s)=(1/s)Ci(s)sin(s)+((π/2)−Si(s))((cos(s))/s) (info.... Ci(z)=∫ ((cos(z))/z)dz , Si(z)=∫ ((sin(z))/z)dz) A→ F(ms)=∫_0 ^∞ tan^(−1) (t)e^(−mst) dt mt=u (du/dt)=m → (1/m)du=dt ∫_0 ^∞ tan^(−1) ((u/m))e^(−ut) du =(1/t)Ci(mt)sin(mt)+((π/2)−Si(mt))((cos(mt))/t) ∴ ∫_0 ^∞ tan^(−1) ((u/m))e^(−u) du= Ci(m)sin(m)+((π/2)−Si(m))cos(m) lim_(m→∞) {Ci(m)sin(m)+((π/2)−Si(m))cos(m)}=L L=0
$$\mathrm{let}'\mathrm{s}\:\mathrm{consider}\:{F}\left({s}\right)=\int_{\mathrm{0}} ^{\infty\:} \:\mathrm{tan}^{−\mathrm{1}} \left({t}\right){e}^{−{st}} \mathrm{d}{t}\:\centerdot\centerdot\centerdot\centerdot\boldsymbol{\mathrm{A}} \\ $$$$\boldsymbol{\mathcal{L}}_{{s}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}} \centerdot\:\:\:\left(\mathrm{Laplace}\:\mathrm{Transform}\right) \\ $$$${F}\left({s}\right)=\frac{\mathrm{1}}{{s}}\mathrm{Ci}\left({s}\right)\mathrm{sin}\left({s}\right)+\left(\frac{\pi}{\mathrm{2}}−\mathrm{Si}\left({s}\right)\right)\frac{\mathrm{cos}\left({s}\right)}{{s}} \\ $$$$\left(\mathrm{info}….\:\mathrm{Ci}\left({z}\right)=\int\:\:\frac{\mathrm{cos}\left({z}\right)}{{z}}\mathrm{d}{z}\:,\:\mathrm{Si}\left({z}\right)=\int\:\:\frac{\mathrm{sin}\left({z}\right)}{{z}}\mathrm{d}{z}\right) \\ $$$$\boldsymbol{\mathrm{A}}\rightarrow \\ $$$${F}\left({ms}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{tan}^{−\mathrm{1}} \left({t}\right){e}^{−{mst}} \mathrm{d}{t} \\ $$$${mt}={u} \\ $$$$\frac{\mathrm{d}{u}}{\mathrm{d}{t}}={m}\:\:\:\rightarrow\:\frac{\mathrm{1}}{{m}}\mathrm{d}{u}=\mathrm{d}{t} \\ $$$$\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{{m}}\right){e}^{−{ut}} \mathrm{d}{u} \\ $$$$=\frac{\mathrm{1}}{{t}}\mathrm{Ci}\left({mt}\right)\mathrm{sin}\left({mt}\right)+\left(\frac{\pi}{\mathrm{2}}−\mathrm{Si}\left({mt}\right)\right)\frac{\mathrm{cos}\left({mt}\right)}{{t}} \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{{m}}\right){e}^{−{u}} \mathrm{d}{u}= \\ $$$$\mathrm{Ci}\left({m}\right)\mathrm{sin}\left({m}\right)+\left(\frac{\pi}{\mathrm{2}}−\mathrm{Si}\left({m}\right)\right)\mathrm{cos}\left({m}\right) \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\:\left\{\mathrm{Ci}\left({m}\right)\mathrm{sin}\left({m}\right)+\left(\frac{\pi}{\mathrm{2}}−\mathrm{Si}\left({m}\right)\right)\mathrm{cos}\left({m}\right)\right\}=\boldsymbol{\mathrm{L}} \\ $$$$\mathrm{L}=\mathrm{0} \\ $$
Answered by Berbere last updated on 30/Mar/24
U_n (x)=tan^(−1) ((x/n))e^(−x) ≤(π/2)e^(−x) ;∀x∈R_+ x→(π/2)e^(−x) ;is Riemann integrabl over [0,∞[ ∫_0 ^∞ tan^(−1) ((x/n))e^(−x) dx≤∫_0 ^∞ (π/2)e^(−x) =(π/2) domiate cv[Theorem⇒ lim_(n→∞) ∫_0 ^∞ tan^(−1) ((x/n))e^(−x) dx=∫_0 ^∞ lim_(n→∞) tan^(−1) ((x/n))e^(−x) dx=0 or using 0≤tan^(−1) (x)≤x tan^(−1) (x)=∫_0 ^x (dt/(1+t^2 ))≤∫_0 ^x dt=x ⇒0≤tan^(−1) ((x/n))e^(−x) ≤(x/n)e^(−x) 0≤∫_0 ^∞ tan^(−1) ((x/n))e^(−x) ≤(1/n)∫_0 ^∞ xe^(−x) dx=(1/n)Γ(2)_(→0) lim_(n→∞) ∫_0 ^∞ tan^(−1) ((x/n))e^(−x) dx=0
$${U}_{{n}} \left({x}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right){e}^{−{x}} \leqslant\frac{\pi}{\mathrm{2}}{e}^{−{x}} ;\forall{x}\in\mathbb{R}_{+} \\ $$$${x}\rightarrow\frac{\pi}{\mathrm{2}}{e}^{−{x}} ;{is}\:{Riemann}\:{integrabl}\:{over}\:\left[\mathrm{0},\infty\left[\right.\right. \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx}\leqslant\int_{\mathrm{0}} ^{\infty} \frac{\pi}{\mathrm{2}}{e}^{−{x}} =\frac{\pi}{\mathrm{2}} \\ $$$${domiate}\:{cv}\left[{Theorem}\Rightarrow\right. \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\infty} \mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx}=\int_{\mathrm{0}} ^{\infty} \underset{{n}\rightarrow\infty} {\mathrm{lim}tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx}=\mathrm{0} \\ $$$${or}\:{using}\:\mathrm{0}\leqslant\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\leqslant{x} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\leqslant\int_{\mathrm{0}} ^{{x}} {dt}={x} \\ $$$$\Rightarrow\mathrm{0}\leqslant\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right){e}^{−{x}} \leqslant\frac{{x}}{{n}}{e}^{−{x}} \\ $$$$\mathrm{0}\leqslant\int_{\mathrm{0}} ^{\infty} \mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right){e}^{−{x}} \leqslant\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}} {dx}=\frac{\mathrm{1}}{{n}}\Gamma\left(\mathrm{2}\right)_{\rightarrow\mathrm{0}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\infty} \mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx}=\mathrm{0} \\ $$$$ \\ $$
Commented by MathedUp last updated on 30/Mar/24
Goooooood
$$\mathrm{Goooooood} \\ $$

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