laplace transform... L { sin((√t) )} =? −−−−


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Question Number 205599 by mnjuly1970 last updated on 25/Mar/24

$l a p l a c e t r a n s f o r m . . .$ $L {s i n (\sqrt{t})} = ?$ $- - - -$
Commented by SANOGO last updated on 25/Mar/24

$m e r c i b e a u c o u p$
	Answered by Berbere last updated on 25/Mar/24
	$L{sin((√t))}(x)=∫_0 ^∞ sin((√t))e^(−xt) dt;x>0 (√t)=y =∫_0 ^∞ sin(y)e^(−xy^2 ) 2ydy=^(IBP) [−(e^(−xy^2 ) /x)sin(y)]_0 ^∞ +(1/x)∫_0 ^∞ cos(y)e^(−xy^2 ) dy =(1/x)Re∫_0 ^∞ e^(−xy^2 +iy) dy=(1/x)Re∫_0 ^∞ e^(−(y(√x)+(i/(2(√x))))^2 −(1/(4x))) dy y(√x).=u,(e^(−(1/(4x))) /(x(√x)))Re∫_0 ^∞ e^(−(u+(i/(2(√x))))^2 ) du X=u+(i/(2(√x)));(e^(−(1/(4x))) /(x(√x)))Re∫_(0+(i/(2(√x)))) ^(∞+(i/(2(√x)))) e^(−X^2 ) dX; let rectangle ofDq (0,0) (0,(i/(2(√x))));(∞,(i/(2(√x))));(∞,0) f(z)=e^(−z^2 ) ∫_D f(z)dz=0 Holomorphic function cauchy Theorem ∫_0 ^(i/(2(√x))) f(z)dz+∫_(i/(2(√x))) ^(∞+(i/(2(√x)))) f(z)dz+∫_(∞+(i/(2(√x)))) ^∞ f(z)dz_(=0) +∫_∞ ^0 f(z)dz=0 z=x+iy ∣f(z)∣≤e^(−y^2 ) .e^(−x^2 ) →0 x→∞ ∫_0 ^∞ e^(−z^2 ) =(1/2)Γ((1/2))=((√π)/2);∫_0 ^(i/(2(√x))) e^(−z^2 ) dz;z=it =i∫_0 ^(1/(2(√x))) e^t^2 dt imaginair Pur Re∫_(0+(i/(2(√x)))) ^(∞+(i/(2(√x)))) e^(−X^2 ) =((√π)/2) L(sin((√t))(x)=(e^(−(1/(4x))) /(x(√x))).((√π)/2)$
	$L {s i n (\sqrt{t})} (x) = \int_{0}^{\infty} s i n (\sqrt{t}) e^{- x t} d t; x > 0$ $\sqrt{t} = y$ $= \int_{0}^{\infty} s i n (y) e^{- {x y}^{2}} 2 y d y \overset{I B P}{=} {[- \frac{e^{- {x y}^{2}}}{x} s i n (y)]}_{0}^{\infty} + \frac{1}{x} \int_{0}^{\infty} c o s (y) e^{- {x y}^{2}} d y$ $= \frac{1}{x} R e \int_{0}^{\infty} e^{- {x y}^{2} + i y} d y = \frac{1}{x} R e \int_{0}^{\infty} e^{- {(y \sqrt{x} + \frac{i}{2 \sqrt{x}})}^{2} - \frac{1}{4 x}} d y$ $y \sqrt{x} . = u, \frac{e^{- \frac{1}{4 x}}}{x \sqrt{x}} R e \int_{0}^{\infty} e^{- {(u + \frac{i}{2 \sqrt{x}})}^{2}} d u$ $X = u + \frac{i}{2 \sqrt{x}}; \frac{e^{- \frac{1}{4 x}}}{x \sqrt{x}} R e \int_{0 + \frac{i}{2 \sqrt{x}}}^{\infty + \frac{i}{2 \sqrt{x}}} e^{- X^{2}} d X;$ $l e t r e c t a n g l e o f D q (0, 0) (0, \frac{i}{2 \sqrt{x}}); (\infty, \frac{i}{2 \sqrt{x}}); (\infty, 0)$ $f (z) = e^{- z^{2}}$ $\int_{D} f (z) d z = 0 H o l o m o r p h i c f u n c t i o n c a u c h y T h e o r e m$ $\int_{0}^{\frac{i}{2 \sqrt{x}}} f (z) d z + \int_{\frac{i}{2 \sqrt{x}}}^{\infty + \frac{i}{2 \sqrt{x}}} f (z) d z + \int_{\infty + \frac{i}{2 \sqrt{x}}}^{\infty} f (z) {d z}_{= 0} + \int_{\infty}^{0} f (z) d z = 0$ $z = x + i y ∣ f (z) ∣⩽ e^{- y^{2}} . e^{- x^{2}} \to 0 x \to \infty$ $\int_{0}^{\infty} e^{- z^{2}} = \frac{1}{2} Γ (\frac{1}{2}) = \frac{\sqrt{π}}{2}; \int_{0}^{\frac{i}{2 \sqrt{x}}} e^{- z^{2}} d z; z = i t$ $= i \int_{0}^{\frac{1}{2 \sqrt{x}}} e^{t^{2}} d t i m a g i n a i r P u r$ $R e \int_{0 + \frac{i}{2 \sqrt{x}}}^{\infty + \frac{i}{2 \sqrt{x}}} e^{- X^{2}} = \frac{\sqrt{π}}{2}$ $L (s i n (\sqrt{t}) (x) = \frac{e^{- \frac{1}{4 x}}}{x \sqrt{x}} . \frac{\sqrt{π}}{2}$
	Commented by mnjuly1970 last updated on 25/Mar/24

	$t h a n k s a l o t M a s t e r$
	Commented by Berbere last updated on 27/Mar/24

	$w i t h e p l e a s u r$
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