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Question Number 130200 by mnjuly1970 last updated on 23/Jan/21

$. . . n i c e c s l c u l u s . . .$ $f i n d :: L ({t e}^{- t} l n (t)) \underset{t r a n s f o r m}{\overset{l a p l a c e}{=}} ?$
	Answered by Dwaipayan Shikari last updated on 23/Jan/21

	$\int_{0}^{\infty} e^{- (s + 1) t} t l o g (t) d t (s + 1) u$ $= \frac{1}{{(s + 1)}^{2}} \int_{0}^{\infty} e^{- u} u l o g (u) - \frac{l o g (s + 1)}{{(s + 1)}^{2}} \int_{0}^{\infty} {u e}^{- u}$ $= \frac{ψ (2) - l o g (1 + s)}{{(s + 1)}^{2}} = - \frac{γ + 1 + l o g (1 + s)}{{(s + 1)}^{2}}$
	Commented by mnjuly1970 last updated on 23/Jan/21

	$t h a n k y o u$ $m e r c e y . . .$
	Answered by mathmax by abdo last updated on 23/Jan/21
	$L(te^(−t) lnt)=∫_0 ^∞ xe^(−x) lnx e^(−tx) dx (t>0) =∫_0 ^∞ x e^(−(t+1)x) lnxdx =_((t+1)x=u) ∫_0 ^∞ (u/(t+1))e^(−u) ln((u/(t+1)))(du/(t+1)) =(1/((t+1)^2 ))∫_0 ^∞ ue^(−u) (lnu−ln(t+1))du =(1/((t+1)^2 )){∫_0 ^∞ ue^(−u) lnudu−ln(t+1)∫_0 ^∞ ue^(−u) du} ∫_0 ^∞ u e^(−u) du =Γ(2)=1 Γ(x)=∫_0 ^∞ u^(x−1) e^(−u) du =∫_0 ^∞ e^((x−1)lnu) e^(−u) du ⇒Γ^′ (x)=∫_0 ^∞ e^(−u) lnu u^(x−1) du ⇒∫_0 ^∞ ue^(−u) lnudu=Γ^′ (2) ⇒ L(te^(−t) lnt)=((Γ^′ (2))/((t+1)^2 ))−((ln(t+1))/((t+1)^2 ))$
	$L ({te}^{- t} lnt) = \int_{0}^{\infty} {xe}^{- x} lnx e^{- tx} dx (t > 0)$ $= \int_{0}^{\infty} x e^{- (t + 1) x} lnxdx =_{(t + 1) x = u} \int_{0}^{\infty} \frac{u}{t + 1} e^{- u} \ln (\frac{u}{t + 1}) \frac{du}{t + 1}$ $= \frac{1}{{(t + 1)}^{2}} \int_{0}^{\infty} {ue}^{- u} (lnu - \ln (t + 1)) du$ $= \frac{1}{{(t + 1)}^{2}} {\int_{0}^{\infty} {ue}^{- u} lnudu - \ln (t + 1) \int_{0}^{\infty} {ue}^{- u} du}$ $\int_{0}^{\infty} u e^{- u} du = Γ (2) = 1$ $Γ (x) = \int_{0}^{\infty} u^{x - 1} e^{- u} du = \int_{0}^{\infty} e^{(x - 1) lnu} e^{- u} du$ $\Rightarrow Γ^{^{'}} (x) = \int_{0}^{\infty} e^{- u} lnu u^{x - 1} du \Rightarrow \int_{0}^{\infty} {ue}^{- u} lnudu = Γ^{^{'}} (2) \Rightarrow$ $L ({te}^{- t} lnt) = \frac{Γ^{^{'}} (2)}{{(t + 1)}^{2}} - \frac{\ln (t + 1)}{{(t + 1)}^{2}}$
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