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 200155 by Calculusboy last updated on 15/Nov/23

Answered by Frix last updated on 15/Nov/23

s>0  ∫_0 ^∞ (√t)e^(−st) dt=  =−(1/s)[(√t)e^(−st) ]_0 ^∞ +(1/(2s))×∫_0 ^∞ (e^(−st) /( (√t)))dt  −(1/s)[(√t)e^(−st) ]_0 ^∞ =0  (1/(2s))t×∫_0 ^∞ (e^(−st) /( (√t)))dt =^(u=(√(st)))  s^(−(3/2)) ×∫_0 ^∞ e^(−u^2 ) du=((√(πs^(−3) ))/2)

$${s}>\mathrm{0} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\sqrt{{t}}\mathrm{e}^{−{st}} {dt}= \\ $$$$=−\frac{\mathrm{1}}{{s}}\left[\sqrt{{t}}\mathrm{e}^{−{st}} \right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}{s}}×\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{e}^{−{st}} }{\:\sqrt{{t}}}{dt} \\ $$$$−\frac{\mathrm{1}}{{s}}\left[\sqrt{{t}}\mathrm{e}^{−{st}} \right]_{\mathrm{0}} ^{\infty} =\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{s}}{t}×\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{e}^{−{st}} }{\:\sqrt{{t}}}{dt}\:\overset{{u}=\sqrt{{st}}} {=}\:{s}^{−\frac{\mathrm{3}}{\mathrm{2}}} ×\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{u}^{\mathrm{2}} } {du}=\frac{\sqrt{\pi{s}^{−\mathrm{3}} }}{\mathrm{2}} \\ $$

Commented by Calculusboy last updated on 15/Nov/23

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

Answered by Mathspace last updated on 15/Nov/23

I=∫_0 ^∞  (√t)e^(−st) dt =_(st=x)   ∫_0 ^∞   (√(x/s))e^(−x) (dx/s)  =(1/s^(3/2) )∫_0 ^∞   x^(1/2)  e^(−x) dx  =(1/s^(3/2) )×Γ((3/2))=(1/s^(3/2) )(1/2)Γ((1/2))  =((√π)/(2s^(3/2) ))

$${I}=\int_{\mathrm{0}} ^{\infty} \:\sqrt{{t}}{e}^{−{st}} {dt}\:=_{{st}={x}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\sqrt{\frac{{x}}{{s}}}{e}^{−{x}} \frac{{dx}}{{s}} \\ $$$$=\frac{\mathrm{1}}{{s}^{\frac{\mathrm{3}}{\mathrm{2}}} }\int_{\mathrm{0}} ^{\infty} \:\:{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−{x}} {dx} \\ $$$$=\frac{\mathrm{1}}{{s}^{\frac{\mathrm{3}}{\mathrm{2}}} }×\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{{s}^{\frac{\mathrm{3}}{\mathrm{2}}} }\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}{s}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$

Commented by Calculusboy last updated on 16/Nov/23

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com