Question Number 206730 by mathzup last updated on 23/Apr/24
$${find}\:{lim}_{{n}\rightarrow+\infty} \int_{\mathrm{0}} ^{{n}} {e}^{{nx}} \:{arctan}\left(\frac{{x}}{{n}}\right){dx} \\ $$
Commented by aleks041103 last updated on 24/Apr/24
![More interesting is lim_(n→∞) ∫_0 ^( n) e^(−nx) arctan((x/n))dx = lim_(n→∞) J_n J_n = ∫_0 ^( n) e^(−nx) arctan((x/n))dx = = ∫_0 ^( 1) ne^(−n^2 t) arctan(t)dt = ∫_0 ^( 1) f_n (t)dt obviously if we fix t then: lim_(n→∞) f_n (t) = arctan(t)lim_(n→∞) ne^(−n^2 t) = 0 lim_(n→∞) f_n (t)=0, t∈[0,1] ⇒f_n →^([0,1]) 0 pointwise but also: arctan(t)≤t for t∈[0,1] ⇒0≤f_n (t)≤nte^(−n^2 t) =(1/n) (ae^(−a) ), a=n^2 t∈[0,n^2 ]⊂[0,∞) ⇒f_n (t)≤(1/n)(ae^(−a) )≤(1/n)(sup_(a∈[0,n^2 ]) ae^(−a) )≤(1/n)(sup_(a≥0) ae^(−a) ) but g(a)=ae^(−a) ⇒g′=(1−a)e^(−a) =0⇒a=1 ⇒sup_(a≥0) ae^(−a) = g(1) = e^(−1) ⇒0≤f_n (t)≤(1/(en)), t∈[0,1] ⇒∥f_n ∥=sup_(t∈[0,1]) ∣f_n (t)∣≤(1/(en)) ⇒lim_(n→∞) ∥f_n ∥=0 ⇒f_n ⇉^([0,1]) 0, i.e. uniformly ⇒lim_(n→∞) ∫_0 ^( 1) f_n (t)dt = ∫_0 ^( 1) [lim_(n→∞) f_n (t)]dt= =∫_0 ^( 1) 0dt=0 ⇒lim_(n→∞) J_n =0 ⇒lim_(n→∞) ∫_0 ^( n) e^(−nx) arctan((x/n))dx = 0](https://www.tinkutara.com/question/Q206778.png)
$${More}\:{interesting}\:{is} \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\:\int_{\mathrm{0}} ^{\:{n}} {e}^{−{nx}} {arctan}\left(\frac{{x}}{{n}}\right){dx}\:=\:\underset{{n}\rightarrow\infty} {{lim}}\:{J}_{{n}} \\ $$$$ \\ $$$${J}_{{n}} =\:\int_{\mathrm{0}} ^{\:{n}} {e}^{−{nx}} {arctan}\left(\frac{{x}}{{n}}\right){dx}\:=\: \\ $$$$=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ne}^{−{n}^{\mathrm{2}} {t}} {arctan}\left({t}\right){dt}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}_{{n}} \left({t}\right){dt} \\ $$$${obviously}\:{if}\:{we}\:{fix}\:{t}\:{then}: \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\:{f}_{{n}} \left({t}\right)\:=\:{arctan}\left({t}\right)\underset{{n}\rightarrow\infty} {{lim}}\:{ne}^{−{n}^{\mathrm{2}} {t}} \:=\:\mathrm{0}\: \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\:{f}_{{n}} \left({t}\right)=\mathrm{0},\:{t}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\Rightarrow{f}_{{n}} \overset{\left[\mathrm{0},\mathrm{1}\right]} {\rightarrow}\:\mathrm{0}\:{pointwise} \\ $$$$ \\ $$$${but}\:{also}: \\ $$$${arctan}\left({t}\right)\leqslant{t}\:{for}\:{t}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\Rightarrow\mathrm{0}\leqslant{f}_{{n}} \left({t}\right)\leqslant{nte}^{−{n}^{\mathrm{2}} {t}} =\frac{\mathrm{1}}{{n}}\:\left({ae}^{−{a}} \right),\:{a}={n}^{\mathrm{2}} {t}\in\left[\mathrm{0},{n}^{\mathrm{2}} \right]\subset\left[\mathrm{0},\infty\right) \\ $$$$\Rightarrow{f}_{{n}} \left({t}\right)\leqslant\frac{\mathrm{1}}{{n}}\left({ae}^{−{a}} \right)\leqslant\frac{\mathrm{1}}{{n}}\left(\underset{{a}\in\left[\mathrm{0},{n}^{\mathrm{2}} \right]} {{sup}}\:{ae}^{−{a}} \right)\leqslant\frac{\mathrm{1}}{{n}}\left(\underset{{a}\geqslant\mathrm{0}} {{sup}}\:{ae}^{−{a}} \right) \\ $$$${but} \\ $$$${g}\left({a}\right)={ae}^{−{a}} \Rightarrow{g}'=\left(\mathrm{1}−{a}\right){e}^{−{a}} =\mathrm{0}\Rightarrow{a}=\mathrm{1} \\ $$$$\Rightarrow\underset{{a}\geqslant\mathrm{0}} {{sup}}\:{ae}^{−{a}} \:=\:{g}\left(\mathrm{1}\right)\:=\:{e}^{−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{0}\leqslant{f}_{{n}} \left({t}\right)\leqslant\frac{\mathrm{1}}{{en}},\:{t}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\Rightarrow\parallel{f}_{{n}} \parallel=\underset{{t}\in\left[\mathrm{0},\mathrm{1}\right]} {{sup}}\:\mid{f}_{{n}} \left({t}\right)\mid\leqslant\frac{\mathrm{1}}{{en}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\:\parallel{f}_{{n}} \parallel=\mathrm{0} \\ $$$$\Rightarrow{f}_{{n}} \overset{\left[\mathrm{0},\mathrm{1}\right]} {\rightrightarrows}\mathrm{0},\:{i}.{e}.\:{uniformly} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}_{{n}} \left({t}\right){dt}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left[\underset{{n}\rightarrow\infty} {{lim}}\:{f}_{{n}} \left({t}\right)\right]{dt}= \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{0}{dt}=\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\:{J}_{{n}} =\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\:\int_{\mathrm{0}} ^{\:{n}} {e}^{−{nx}} {arctan}\left(\frac{{x}}{{n}}\right){dx}\:=\:\mathrm{0} \\ $$
Answered by aleks041103 last updated on 24/Apr/24
![x=nt⇒dx=ndt ⇒∫_0 ^( n) e^(nx) arctan(x/n)dx=∫_0 ^( 1) e^(n^2 t) arctan(t)ndt=I_n but e^(n^2 t) ≥1 for t∈[0,1] I_n =n∫_0 ^( 1) e^(n^2 t) arctan(t)dt≥n∫_0 ^( 1) arctan(t)dt= =n{[t arctan(t)]_0 ^1 −∫_0 ^( 1) ((tdt)/(1+t^2 ))}= =n{(π/4)−(1/2)[ln(1+t^2 )]_0 ^1 }=n((π/4)−((ln(2))/2)) but (π/4)>(3/4)=0.75 and (1/2)ln(2)<(1/2)=0.5 ⇒c=(π/4)−((ln(2))/2)>0.25>0 ⇒I_n >cn, c>0 ⇒since lim_(n→∞) cn = +∞ then lim_(n→∞) I_n = lim_(n→∞) ∫_0 ^( n) e^(nx) arctan((x/n))dx = +∞](https://www.tinkutara.com/question/Q206777.png)
$${x}={nt}\Rightarrow{dx}={ndt} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:{n}} {e}^{{nx}} {arctan}\left({x}/{n}\right){dx}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {e}^{{n}^{\mathrm{2}} {t}} {arctan}\left({t}\right){ndt}={I}_{{n}} \\ $$$${but}\:{e}^{{n}^{\mathrm{2}} {t}} \geqslant\mathrm{1}\:{for}\:{t}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${I}_{{n}} ={n}\int_{\mathrm{0}} ^{\:\mathrm{1}} {e}^{{n}^{\mathrm{2}} {t}} {arctan}\left({t}\right){dt}\geqslant{n}\int_{\mathrm{0}} ^{\:\mathrm{1}} {arctan}\left({t}\right){dt}= \\ $$$$={n}\left\{\left[{t}\:{arctan}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{tdt}}{\mathrm{1}+{t}^{\mathrm{2}} }\right\}= \\ $$$$={n}\left\{\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \right\}={n}\left(\frac{\pi}{\mathrm{4}}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\right) \\ $$$${but}\:\frac{\pi}{\mathrm{4}}>\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0}.\mathrm{75}\:{and}\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)<\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}.\mathrm{5} \\ $$$$\Rightarrow{c}=\frac{\pi}{\mathrm{4}}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}>\mathrm{0}.\mathrm{25}>\mathrm{0} \\ $$$$\Rightarrow{I}_{{n}} >{cn},\:{c}>\mathrm{0} \\ $$$$\Rightarrow{since}\:\underset{{n}\rightarrow\infty} {{lim}}\:{cn}\:=\:+\infty\:{then} \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\:{I}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {{lim}}\:\int_{\mathrm{0}} ^{\:{n}} {e}^{{nx}} \:{arctan}\left(\frac{{x}}{{n}}\right){dx}\:=\:+\infty \\ $$