... laplace transformation.. L (te^(−t) ⌊t⌋)=? note : ⌊x⌋ is floor of ′′ x ′′... ..............


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Question Number 129318 by mnjuly1970 last updated on 14/Jan/21

$. . . l a p l a c e t r a n s f o r m a t i o n . .$ $L ({t e}^{- t} ⌊ t ⌋) = ?$ $n o t e : ⌊ x ⌋ i s f l o o r o f^{″} x^{″} . . .$ $. . . . . . . . . . . . . .$
	Answered by mathmax by abdo last updated on 14/Jan/21
	$L(te^(−t) [t]) =∫_0 ^∞ xe^(−x) [x] e^(−tx) dx =∫_0 ^∞ x[x]e^(−(t+1)x) dx =Σ_(n=0) ^∞ ∫_n ^(n+1) nxe^(−(t+1)x) dx =Σ_(n=0) ^∞ n ∫_n ^(n+1) xe^(−(t+1)x) dx we have A_n =∫_n ^(n+1) xe^(−(t+1)x) dx =_((t+1)x=u) ∫_((t+1)n) ^((t+1)(n+1)) (u/(t+1))e^(−u) (du/(t+1)) =(1/((t+1)^2 ))∫_(n(t+1)) ^((n+1)(t+1)) ue^(−u) du =(1/((t+1)^2 )){[−ue^(−u) ]_(n(t+1)) ^((n+1)(t+1)) +∫_(n(t+1)) ^((n+1)(t+1)) e^(−u) du} =(1/((t+1)^2 )){n(t+1)e^(−n(t+1)) −(n+1)(t+1)e^(−(n+1)(t+1)) } +(1/((t+1)^2 ))[−e^(−u) ]_(n(t+1)) ^((n+1)(t+1)) =(n/(t+1))e^(−n(t+1)) −((n+1)/(t+1))e^(−(n+1)(t+1)) +(1/((t+1)^2 ))e^(−n(t+1)) −(1/((t+1)^2 ))e^(−(n+1)(t+1)) =((n/(t+1))+(1/((t+1)^2 )))e^(−n(t+1)) −(((n+1)/(t+1))+(1/((t+1)^2 )))e^(−(n+1)(t+1)) ⇒ L(te^(−t) [t]) =Σ_(n=0) ^∞ ((n/(t+1))+(1/((t+1)^2 )))e^(−n(t+1)) −Σ_(n=0) ^∞ (((n+1)/(t+1))+(1/((t+1)^2 )))e^(−(n+1)(t+1)) =S_1 −S_2 S_1 =(1/(t+1))Σ_(n=0) ^∞ n(e^(−(t+1)) )^n +(1/((t+1)^2 ))×(1/(1−e^(−(t+1)) )) =(1/(t+1))Ψ(e^(−(t+1)) )+(1/((t+1)^2 (1−e^(−(t+1)) ))) with Ψ(x)= Σ_(n=0) ^∞ nx^n we have Σ_(n=0) ^∞ x^n =(1/(1−x)) for ∣x∣<1 ⇒ Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒Σ_(n=1) ^∞ nx^n =(x/((1−x)^2 )) ⇒ S_1 =(1/(t+1))×(e^(−(t+1)) /((1−e^(−(t+1)) )^2 ))+(1/((t+1)^2 (1−e^(−(t+1)) ))) S_2 =(1/(t+1))Σ_(n=0) ^∞ (n+1)e^(−(n+1)(t+1)) −(1/((t+1)^2 ))Σ_(n=0) ^∞ e^(−(n+1)(t+1)) =(1/(t+1))Σ_(n=1) ^∞ ne^(−n(t+1)) −(1/((t+1)^2 ))Σ_(n=1) ^∞ e^(−n(t+1)) Σ_(n=1) ^∞ e^(−n(t+1)) =Σ_(n=1) ^∞ (e^(−(t+1)) )^n =(1/(1−e^(−(t+1)) ))−1 =(e^(−(t+1)) /(1−e^(−(t+1)) )) Σ_(n=1) ^∞ ne^(−n(t+1)) =Σ_(n=1) ^∞ n (e^(−(t+1)) )^n =Φ(e^(−(t+1)) ) Φ is know....$
	$L ({te}^{- t} [t]) = \int_{0}^{\infty} {xe}^{- x} [x] e^{- tx} dx = \int_{0}^{\infty} x [x] e^{- (t + 1) x} dx$ $= \sum_{n = 0}^{\infty} \int_{n}^{n + 1} {nxe}^{- (t + 1) x} dx = \sum_{n = 0}^{\infty} n \int_{n}^{n + 1} {xe}^{- (t + 1) x} dx$ $we have A_{n} = \int_{n}^{n + 1} {xe}^{- (t + 1) x} dx =_{(t + 1) x = u} \int_{(t + 1) n}^{(t + 1) (n + 1)} \frac{u}{t + 1} e^{- u} \frac{du}{t + 1}$ $= \frac{1}{{(t + 1)}^{2}} \int_{n (t + 1)}^{(n + 1) (t + 1)} {ue}^{- u} du = \frac{1}{{(t + 1)}^{2}} {{[- {ue}^{- u}]}_{n (t + 1)}^{(n + 1) (t + 1)}$ $+ \int_{n (t + 1)}^{(n + 1) (t + 1)} e^{- u} du}$ $= \frac{1}{{(t + 1)}^{2}} {n (t + 1) e^{- n (t + 1)} - (n + 1) (t + 1) e^{- (n + 1) (t + 1)}}$ $+ \frac{1}{{(t + 1)}^{2}} {[- e^{- u}]}_{n (t + 1)}^{(n + 1) (t + 1)}$ $= \frac{n}{t + 1} e^{- n (t + 1)} - \frac{n + 1}{t + 1} e^{- (n + 1) (t + 1)} + \frac{1}{{(t + 1)}^{2}} e^{- n (t + 1)} - \frac{1}{{(t + 1)}^{2}} e^{- (n + 1) (t + 1)}$ $= (\frac{n}{t + 1} + \frac{1}{{(t + 1)}^{2}}) e^{- n (t + 1)} - (\frac{n + 1}{t + 1} + \frac{1}{{(t + 1)}^{2}}) e^{- (n + 1) (t + 1)} \Rightarrow$ $L ({te}^{- t} [t]) = \sum_{n = 0}^{\infty} (\frac{n}{t + 1} + \frac{1}{{(t + 1)}^{2}}) e^{- n (t + 1)}$ $- \sum_{n = 0}^{\infty} (\frac{n + 1}{t + 1} + \frac{1}{{(t + 1)}^{2}}) e^{- (n + 1) (t + 1)} = S_{1} - S_{2}$ $S_{1} = \frac{1}{t + 1} \sum_{n = 0}^{\infty} n {(e^{- (t + 1)})}^{n} + \frac{1}{{(t + 1)}^{2}} \times \frac{1}{1 - e^{- (t + 1)}} = \frac{1}{t + 1} Ψ (e^{- (t + 1)}) + \frac{1}{{(t + 1)}^{2} (1 - e^{- (t + 1)})} with Ψ (x) =$ $\sum_{n = 0}^{\infty} {nx}^{n} we have \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} for ∣ x ∣< 1 \Rightarrow$ $\sum_{n = 1}^{\infty} {nx}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow \sum_{n = 1}^{\infty} {nx}^{n} = \frac{x}{{(1 - x)}^{2}} \Rightarrow$ $S_{1} = \frac{1}{t + 1} \times \frac{e^{- (t + 1)}}{{(1 - e^{- (t + 1)})}^{2}} + \frac{1}{{(t + 1)}^{2} (1 - e^{- (t + 1)})}$ $S_{2} = \frac{1}{t + 1} \sum_{n = 0}^{\infty} (n + 1) e^{- (n + 1) (t + 1)} - \frac{1}{{(t + 1)}^{2}} \sum_{n = 0}^{\infty} e^{- (n + 1) (t + 1)}$ $= \frac{1}{t + 1} \sum_{n = 1}^{\infty} {ne}^{- n (t + 1)} - \frac{1}{{(t + 1)}^{2}} \sum_{n = 1}^{\infty} e^{- n (t + 1)}$ $\sum_{n = 1}^{\infty} e^{- n (t + 1)} = \sum_{n = 1}^{\infty} {(e^{- (t + 1)})}^{n} = \frac{1}{1 - e^{- (t + 1)}} - 1 = \frac{e^{- (t + 1)}}{1 - e^{- (t + 1)}}$ $\sum_{n = 1}^{\infty} {ne}^{- n (t + 1)} = \sum_{n = 1}^{\infty} n {(e^{- (t + 1)})}^{n} = Φ (e^{- (t + 1)})$ $Φ is know . . . .$
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