Question Number 36186 by prof Abdo imad last updated on 30/May/18
$${find}\:{nature}\:{of}\:\int_{\mathrm{1}} ^{+\infty} \sqrt{{t}}\:{sin}\left({t}^{\mathrm{2}} \right){dt}\:. \\ $$
Commented by maxmathsup by imad last updated on 15/Aug/18
![let I = ∫_1 ^(+∞) (√t)sin(t^2 )dt changement t^2 =u give I = ∫_1 ^(+∞) u^(1/4) sin(u) (1/2)u^((1/2)−1) du = (1/2)∫_1 ^(+∞) u^((3/4)−1) sin(u)du =(1/2) ∫_1 ^(+∞) u^(−(1/4)) sinu du by parts α =u^(−(1/4)) and β^′ =sinu 2I =[ −u^(−(1/4)) cosu]_1 ^(+∞) −∫_1 ^(+∞) −(1/4) u^(−(5/4)) (−cosu)du =cos1 −(1/4) ∫_1 ^(+∞) ((cosu)/u^(5/4) ) du but ∣ ∫_1 ^(+∞) ((cosu)/u^(5/4) )du∣≤∫_1 ^(+∞) (du/u^(5/4) ) and this integral converges because (5/4)>1 ⇒ I converges .](Q41947.png)
$${let}\:{I}\:=\:\int_{\mathrm{1}} ^{+\infty} \:\sqrt{{t}}{sin}\left({t}^{\mathrm{2}} \right){dt}\:\:{changement}\:{t}^{\mathrm{2}} ={u}\:{give} \\ $$$${I}\:\:=\:\int_{\mathrm{1}} ^{+\infty} \:\:{u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{sin}\left({u}\right)\:\frac{\mathrm{1}}{\mathrm{2}}{u}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {du}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{+\infty} \:\:{u}^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} \:{sin}\left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{+\infty} \:\:{u}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{sinu}\:{du}\:\:\:{by}\:{parts}\:\:\alpha\:={u}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\:{and}\:\beta^{'} \:={sinu} \\ $$$$\mathrm{2}{I}\:=\left[\:−{u}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{cosu}\right]_{\mathrm{1}} ^{+\infty} \:−\int_{\mathrm{1}} ^{+\infty} \:\:−\frac{\mathrm{1}}{\mathrm{4}}\:{u}^{−\frac{\mathrm{5}}{\mathrm{4}}} \left(−{cosu}\right){du} \\ $$$$={cos}\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{cosu}}{{u}^{\frac{\mathrm{5}}{\mathrm{4}}} }\:{du}\:\:\:{but}\:\:\:\mid\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{cosu}}{{u}^{\frac{\mathrm{5}}{\mathrm{4}}} }{du}\mid\leqslant\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{du}}{{u}^{\frac{\mathrm{5}}{\mathrm{4}}} }\:\:{and}\:{this}\:{integral} \\ $$$${converges}\:{because}\:\frac{\mathrm{5}}{\mathrm{4}}>\mathrm{1}\:\:\Rightarrow\:{I}\:\:{converges}\:. \\ $$