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)](Q200157.png)
$${s}>\mathrm{0} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\sqrt{{t}}\mathrm{e}^{−{st}} {dt}= \\ $$$$=−\frac{\mathrm{1}}{{s}}\left[\sqrt{{t}}\mathrm{e}^{−{st}} \right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}{s}}×\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{e}^{−{st}} }{\:\sqrt{{t}}}{dt} \\ $$$$−\frac{\mathrm{1}}{{s}}\left[\sqrt{{t}}\mathrm{e}^{−{st}} \right]_{\mathrm{0}} ^{\infty} =\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{s}}{t}×\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{e}^{−{st}} }{\:\sqrt{{t}}}{dt}\:\overset{{u}=\sqrt{{st}}} {=}\:{s}^{−\frac{\mathrm{3}}{\mathrm{2}}} ×\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{u}^{\mathrm{2}} } {du}=\frac{\sqrt{\pi{s}^{−\mathrm{3}} }}{\mathrm{2}} \\ $$
Commented by Calculusboy last updated on 15/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by Mathspace last updated on 15/Nov/23
$${I}=\int_{\mathrm{0}} ^{\infty} \:\sqrt{{t}}{e}^{−{st}} {dt}\:=_{{st}={x}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\sqrt{\frac{{x}}{{s}}}{e}^{−{x}} \frac{{dx}}{{s}} \\ $$$$=\frac{\mathrm{1}}{{s}^{\frac{\mathrm{3}}{\mathrm{2}}} }\int_{\mathrm{0}} ^{\infty} \:\:{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−{x}} {dx} \\ $$$$=\frac{\mathrm{1}}{{s}^{\frac{\mathrm{3}}{\mathrm{2}}} }×\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{{s}^{\frac{\mathrm{3}}{\mathrm{2}}} }\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}{s}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$
Commented by Calculusboy last updated on 16/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$