Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 107107 by mohammad17 last updated on 08/Aug/20

Commented by Dwaipayan Shikari last updated on 08/Aug/20

∫_0 ^∞ (√y)  e^(−t^2 )                    y^3 =t^(2  )  ,y^(3/2) =t     (3/2)(√y)  =(dt/dy)  (2/3)∫_0 ^∞ (3/2)(√y) e^(−y^3 ) dy=(2/3)∫_0 ^∞ e^(−t^2 ) dt=(2/3).((√π)/2)=((√π)/3)

$$\int_{\mathrm{0}} ^{\infty} \sqrt{\mathrm{y}}\:\:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}^{\mathrm{3}} =\mathrm{t}^{\mathrm{2}\:\:} \:,\mathrm{y}^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{t}\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{y}}\:\:=\frac{\mathrm{dt}}{\mathrm{dy}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{y}}\:\mathrm{e}^{−\mathrm{y}^{\mathrm{3}} } \mathrm{dy}=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}=\frac{\mathrm{2}}{\mathrm{3}}.\frac{\sqrt{\pi}}{\mathrm{2}}=\frac{\sqrt{\pi}}{\mathrm{3}} \\ $$

Commented by PRITHWISH SEN 2 last updated on 08/Aug/20

put y^3 = t⇒y=t^(1/3) ⇒dy=(1/3)t^(−(2/3)) dt  =(1/3)∫_0 ^∞  t^((1/2)−1) e^(−t) dt = (1/3)Γ((1/2))=((√π)/3)

$$\mathrm{put}\:\mathrm{y}^{\mathrm{3}} =\:\mathrm{t}\Rightarrow\mathrm{y}=\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}} \Rightarrow\mathrm{dy}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{3}} \\ $$

Answered by mathmax by abdo last updated on 08/Aug/20

let I =∫_0 ^∞  (√y)e^(−y^3 ) dy  we do the changement y^3 =t ⇒y=t^(1/3)  ⇒  I =∫_0 ^∞  t^(1/6)  e^(−t)  (1/3)t^((1/3)−1)  dt =(1/3)∫_0 ^∞  t^((1/6)+(1/3)−1)  e^(−t)  dt  =(1/3)∫_0 ^∞  t^((1/2)−1)  e^(−t)  dt  =(1/3)∫_0 ^∞   (e^(−t) /(√t))dt =_((√t)=u)  (1/3)∫_0 ^∞  (e^(−u^2 ) /u)(2u)du  =(2/3)∫_0 ^∞  e^(−u^2 ) du =(2/3).((√π)/2)=((√π)/3) ⇒I =((√π)/3)

$$\mathrm{let}\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\sqrt{\mathrm{y}}\mathrm{e}^{−\mathrm{y}^{\mathrm{3}} } \mathrm{dy}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{y}^{\mathrm{3}} =\mathrm{t}\:\Rightarrow\mathrm{y}=\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\Rightarrow \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{6}}} \:\mathrm{e}^{−\mathrm{t}} \:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{e}^{−\mathrm{t}} }{\sqrt{\mathrm{t}}}\mathrm{dt}\:=_{\sqrt{\mathrm{t}}=\mathrm{u}} \:\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } }{\mathrm{u}}\left(\mathrm{2u}\right)\mathrm{du} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}\:=\frac{\mathrm{2}}{\mathrm{3}}.\frac{\sqrt{\pi}}{\mathrm{2}}=\frac{\sqrt{\pi}}{\mathrm{3}}\:\Rightarrow\mathrm{I}\:=\frac{\sqrt{\pi}}{\mathrm{3}} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com