Question Number 208318 by SANOGO last updated on 11/Jun/24
$${calcul}\:\:\:{lim}\:{n}\rightarrow+\infty \\ $$$$\int_{\mathrm{0}} ^{+\infty} \:\frac{{cos}\left({nx}\right)}{\left({nx}+\mathrm{1}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:}{dx} \\ $$
Answered by Berbere last updated on 11/Jun/24
![∣∫_0 ^(+∞) ((cos(nx))/((1+nx)(1+x^2 )))dx∣≤∫_0 ^∞ ((∣cos(nx)∣)/((1+nx)(1+x^2 )))dx≤∫_0 ^∞ (dx/((1+nx)(1+x^2 ))) x→(1/x^2 )⇒∫_0 ^∞ ((xdx)/((1+x^2 )(n+x)))=∫_0 ^∞ ((−n)/(1+n^2 )).(1/((n+x)))+((((nx)/(1+n^2 ))+(1/(n^2 +1)))/((1+x^2 )))dx (1/(n^2 +1))∫_0 ^∞ −(n/(n+x))+((nx+1)/(x^2 +1))dx=(1/(n^2 +1))[−nln(n+x)+(n/2)ln(x^2 +1)+tan^(−1) (x)]_0 ^∞ =(n/(n^2 +1))[ln(((√(x^2 +1))/(x+n)))]_0 ^∞ +(π/(2(n^2 +1))) =−((nln(n))/(n^2 +1))+(π/(2(n^2 +1))) ∣∫_0 ^∞ ((cos(nx))/((nx+1)(1+x^2 )))dx∣≤(1/(n^2 +1))((π/2)−nln(n))=(π/(2(n^2 +1)))−(((ln(n))/n)/(1+(1/n^2 )))→^(n→∞) 0](https://www.tinkutara.com/question/Q208319.png)
$$\mid\int_{\mathrm{0}} ^{+\infty} \frac{{cos}\left({nx}\right)}{\left(\mathrm{1}+{nx}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\mid\leqslant\int_{\mathrm{0}} ^{\infty} \frac{\mid{cos}\left({nx}\right)\mid}{\left(\mathrm{1}+{nx}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\leqslant\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{nx}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$${x}\rightarrow\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{xdx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({n}+{x}\right)}=\int_{\mathrm{0}} ^{\infty} \frac{−{n}}{\mathrm{1}+{n}^{\mathrm{2}} }.\frac{\mathrm{1}}{\left({n}+{x}\right)}+\frac{\frac{{nx}}{\mathrm{1}+{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\int_{\mathrm{0}} ^{\infty} −\frac{{n}}{{n}+{x}}+\frac{{nx}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\left[−{nln}\left({n}+{x}\right)+\frac{{n}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{{n}}{{n}^{\mathrm{2}} +\mathrm{1}}\left[{ln}\left(\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+{n}}\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\pi}{\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=−\frac{{nln}\left({n}\right)}{{n}^{\mathrm{2}} +\mathrm{1}}+\frac{\pi}{\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\mid\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({nx}\right)}{\left({nx}+\mathrm{1}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\mid\leqslant\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\left(\frac{\pi}{\mathrm{2}}−{nln}\left({n}\right)\right)=\frac{\pi}{\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\frac{{ln}\left({n}\right)}{{n}}}{\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}\overset{{n}\rightarrow\infty} {\rightarrow}\mathrm{0} \\ $$$$ \\ $$