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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>00testdt==1s[test]0+12s×0esttdt1s[test]0=012st×0esttdt=u=sts32×0eu2du=πs32

Commented by Calculusboy last updated on 15/Nov/23

thanks sir

thankssir

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=0testdt=st=x0xsexdxs=1s320x12exdx=1s32×Γ(32)=1s3212Γ(12)=π2s32

Commented by Calculusboy last updated on 16/Nov/23

thanks sir

thankssir

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