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Question Number 129367 by BHOOPENDRA last updated on 15/Jan/21

Commented by BHOOPENDRA last updated on 15/Jan/21

$u s i n g u n i t s t e p f u n c t i o n f i n d l a p l a c e t r a n s f o r m$
	Answered by mathmax by abdo last updated on 15/Jan/21
	$L(f(t))=∫_0 ^∞ f(x)e^(−xt) dx =∫_0 ^π f(x)e^(−xt) dx+∫_π ^(2π) f(x)e^(−xt) dx+∫_(2π) ^(+∞) f(x)e^(−xt) dx =∫_0 ^π sinx e^(−xt) dx +∫_π ^(2π) sin(2x)e^(−xt) dx+∫_(2π) ^∞ sin(3x)e^(−xt) dx we hsve ∫_0 ^π e^(−xt) sinx dx =Im(∫_0 ^π e^(−xt+ix) dx) a7d ∫_0 ^π e^((−t+i)x) dx =[(1/(−t+i))e^((−t+i)x) ]_0 ^π =−(1/(t−i)){e^((−t+i)π) −1} =−((t+i)/(t^2 +1)){−e^(−πt) −1} =((t+i)/(t^2 +1))(1+e^(−πt) ) ⇒ ∫_0 ^π e^(−tx) sinx dx =((1+e^(−πt) )/(1+t^2 )) ∫_π ^(2π) sin(2x)e^(−tx) dx =Im(∫_π ^(2π) e^(−tx+2ix) dx) and ∫_π ^(2π) e^((−t+2i)x) dx =[(1/(−t+2i))e^((−t+2i)x) ]_π ^(2π) =−(1/(t−2i)){e^((−t+2i)2π) −e^((−t+2i)π) } =−((t+2i)/(t^2 +4)){e^(−2πt) −e^(−πt) } =((t+2i)/(t^2 +4))(e^(−πt) −e^(−2πt) )⇒ ∫_π ^(2π) e^((−t+2)x) dx =(2/(t^2 +4))(e^(−πt) −e^(−2πt) ) ∫_(2π) ^∞ sin(3x)e^(−tx) dx =Im(∫_(2π) ^∞ e^(−tx+3ix) dx) ∫_(2π) ^∞ e^((−t+3i)x) dx =[(1/(−t+3i))e^((−t+3i)x) ]_(2π) ^∞ =−(1/(t−3i)){−e^((−t+3i)2π) } =((t+3i)/(t^2 +9))(e^(−2πt) ) ⇒ ∫_(2π) ^∞ sin(3x)e^(−tx) dx =((3e^(−2πt) )/(t^2 +9)) ⇒ L(f(t)) =((1+e^(−πt) )/(1+t^2 )) +(2/(t^2 +4))(e^(−πt) −e^(−2πt) )+((3e^(−2πt) )/(t^2 +9))$
	$L (f (t)) = \int_{0}^{\infty} f (x) e^{- xt} dx = \int_{0}^{π} f (x) e^{- xt} dx + \int_{π}^{2 π} f (x) e^{- xt} dx + \int_{2 π}^{+ \infty} f (x) e^{- xt} dx$ $= \int_{0}^{π} sinx e^{- xt} dx + \int_{π}^{2 π} \sin (2 x) e^{- xt} dx + \int_{2 π}^{\infty} \sin (3 x) e^{- xt} dx we hsve$ $\int_{0}^{π} e^{- xt} sinx dx = Im (\int_{0}^{π} e^{- xt + ix} dx) a 7 d$ $\int_{0}^{π} e^{(- t + i) x} dx = {[\frac{1}{- t + i} e^{(- t + i) x}]}_{0}^{π} = - \frac{1}{t - i} {e^{(- t + i) π} - 1}$ $= - \frac{t + i}{t^{2} + 1} {- e^{- π t} - 1} = \frac{t + i}{t^{2} + 1} (1 + e^{- π t}) \Rightarrow$ $\int_{0}^{π} e^{- tx} sinx dx = \frac{1 + e^{- π t}}{1 + t^{2}}$ $\int_{π}^{2 π} \sin (2 x) e^{- tx} dx = Im (\int_{π}^{2 π} e^{- tx + 2 ix} dx) and$ $\int_{π}^{2 π} e^{(- t + 2 i) x} dx = {[\frac{1}{- t + 2 i} e^{(- t + 2 i) x}]}_{π}^{2 π}$ $= - \frac{1}{t - 2 i} {e^{(- t + 2 i) 2 π} - e^{(- t + 2 i) π}}$ $= - \frac{t + 2 i}{t^{2} + 4} {e^{- 2 π t} - e^{- π t}} = \frac{t + 2 i}{t^{2} + 4} (e^{- π t} - e^{- 2 π t}) \Rightarrow$ $\int_{π}^{2 π} e^{(- t + 2) x} dx = \frac{2}{t^{2} + 4} (e^{- π t} - e^{- 2 π t})$ $\int_{2 π}^{\infty} \sin (3 x) e^{- tx} dx = Im (\int_{2 π}^{\infty} e^{- tx + 3 ix} dx)$ $\int_{2 π}^{\infty} e^{(- t + 3 i) x} dx = {[\frac{1}{- t + 3 i} e^{(- t + 3 i) x}]}_{2 π}^{\infty}$ $= - \frac{1}{t - 3 i} {- e^{(- t + 3 i) 2 π}} = \frac{t + 3 i}{t^{2} + 9} (e^{- 2 π t}) \Rightarrow$ $\int_{2 π}^{\infty} \sin (3 x) e^{- tx} dx = \frac{{3 e}^{- 2 π t}}{t^{2} + 9} \Rightarrow$ $L (f (t)) = \frac{1 + e^{- π t}}{1 + t^{2}} + \frac{2}{t^{2} + 4} (e^{- π t} - e^{- 2 π t}) + \frac{{3 e}^{- 2 π t}}{t^{2} + 9}$
	Commented by BHOOPENDRA last updated on 15/Jan/21

	$t h a n k y o u s i r$
	Commented by mathmax by abdo last updated on 16/Jan/21

	$you are welcome$
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