Question Number 36184 by prof Abdo imad last updated on 30/May/18
$${study}\:{the}\:{convergence}\:{of}\:\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{cos}\left({t}\right)}{\:\sqrt{{t}}}{dt} \\ $$
Commented by maxmathsup by imad last updated on 19/Aug/18
![by parts u=(1/( (√t))) and v^′ =cost ⇒ ∫_1 ^(+∞) ((cost)/( (√t))) dt =[ ((sint)/( (√t)))]_1 ^(+∞) −∫_1 ^(+∞) (−(1/(2t(√t))))sint dt =sin1 + (1/2)∫_1 ^(+∞) ((sint)/t^(3/2) ) dt but ∣ ∫_1 ^(+∞) ((sint)/t^(3/2) )dt∣ ≤ ∫_1 ^(+∞) (dt/t^(3/2) ) and this intevral converges because (3/2)>1 .](https://www.tinkutara.com/question/Q42165.png)
$${by}\:{parts}\:{u}=\frac{\mathrm{1}}{\:\sqrt{{t}}}\:\:{and}\:{v}^{'} ={cost}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{cost}}{\:\sqrt{{t}}}\:{dt}\:=\left[\:\frac{{sint}}{\:\sqrt{{t}}}\right]_{\mathrm{1}} ^{+\infty} \:−\int_{\mathrm{1}} ^{+\infty} \:\:\left(−\frac{\mathrm{1}}{\mathrm{2}{t}\sqrt{{t}}}\right){sint}\:{dt} \\ $$$$={sin}\mathrm{1}\:\:+\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{sint}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{dt}\:\:{but} \\ $$$$\mid\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{sint}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt}\mid\:\leqslant\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dt}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{and}\:{this}\:{intevral}\:{converges}\:{because}\:\frac{\mathrm{3}}{\mathrm{2}}>\mathrm{1}\:. \\ $$$$ \\ $$