Ayşe T. (İzmir, 6-month user) ⭐⭐
Hititbet giriş adresini bulmak bazen çile oluyor! 3 kez bloke edildi, VPN kullanmak zorunda kaldım. 1.200 TL yatırım yaptım ama hoşgeldi bonusu ertesi gün aktif oldu. Destek ekibi hızlı yanıt veriyor ama çözüm üretmede yetersizler. Oyun çeşitliliği iyi ama arayüz eski moda.
#access_issues #customer_support #deposit_bonus #platform_ux
**Me# 4.1: Introduction to the Laplace Transform
- Page ID
- 30704
\\( \
ewcommand{\\vecs}[1]{\\overset { \\scriptstyle \\rightharpoonup} {\\mathbf{#1}} } \\) \\( \
ewcommand{\\vecd}[1]{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}} \\)\\(\
ewcommand{\\id}{\\mathrm{id}}\\) \\( \
ewcommand{\\Span}{\\mathrm{span}}\\) \\( \
ewcommand{\\kernel}{\\mathrm{null}\\,}\\) \\( \
ewcommand{\\range}{\\mathrm{range}\\,}\\) \\( \
ewcommand{\\RealPart}{\\mathrm{Re}}\\) \\( \
ewcommand{\\ImaginaryPart}{\\mathrm{Im}}\\) \\( \
ewcommand{\\Argument}{\\mathrm{Arg}}\\) \\( \
ewcommand{\
orm}[1]{\\| #1 \\|}\\) \\( \
ewcommand{\\inner}[2]{\\langle #1, #2 \\rangle}\\) \\( \
ewcommand{\\Span}{\\mathrm{span}}\\) \\(\
ewcommand{\\id}{\\mathrm{id}}\\) \\( \
ewcommand{\\Span}{\\mathrm{span}}\\) \\( \
ewcommand{\\kernel}{\\mathrm{null}\\,}\\) \\( \
ewcommand{\\range}{\\mathrm{range}\\,}\\) \\( \
ewcommand{\\RealPart}{\\mathrm{Re}}\\) \\( \
ewcommand{\\ImaginaryPart}{\\mathrm{Im}}\\) \\( \
ewcommand{\\Argument}{\\mathrm{Arg}}\\) \\( \
ewcommand{\
orm}[1]{\\| #1 \\|}\\) \\( \
ewcommand{\\inner}[2]{\\langle #1, #2 \\rangle}\\) \\( \
ewcommand{\\Span}{\\mathrm{span}}\\)\\(\
ewcommand{\\AA}{\\unicode[.8,0]{x212B}}\\)
## Definition of the Laplace Transform
To define the Laplace transform, we first recall the definition of an improper integral. If \\(g\\) is integrable over the interval \\([a,T]\\) for every \\(T>a\\), then the improper integral of \\(g\\) over \\([a,\\infty)\\) is defined as
\\[\\label{eq:8.1.1} \\int_{a}^{\\infty}g(t)\\,dt=\\lim_{T\\to\\infty}\\int_{a}^{T}g(t)\\,dt.\\]
We say that the improper integral converges if the limit in Equation \\ref{eq:8.1.1} exists; otherwise, we say that the improper integral diverges or does not exist. Here’s the definition of the Laplace transform of a function \\(f(t)\\).
Let \\(f(t)\\) be defined for \\(t\\ge0\\) and let \\(s\\) be a real number. Then the Laplace transform of \\(f\\) is the function \\(F\\) defined by
\\[\\label{eq:8.1.2} F(s)=\\int_{0}^{\\infty}e^{-st} f(t)\\,dt,\\]
for those values of \\(s\\) for which the improper integral converges.
It is important to keep in mind that the variable of integration in Equation \\ref{eq:8.1.2} is \\(t\\), while \\(s\\) is a parameter independent of \\(t\\). We use \\(t\\) as the independent variable for \\(f\\) because in applications the Laplace transform is usually applied to functions of time.
The Laplace transform can be viewed as an operator \\({\\cal L}\\) that transforms the function \\(f=f(t)\\) into the function \\(F=F(s)\\). Thus, Equation \\ref{eq:8.1.2} can be expressed as
\\[F={\\cal L}(f).\
onumber \\]
The functions \\(f\\) and \\(F\\) form a transform pair, which we’ll sometimes denote by
\\[f(t)\\leftrightarrow F(s).\
onumber\\]
It can be shown that if \\(F(s)\\) is defined for \\(s=s_{0}\\) then it is defined for all \\(s>s_{0}\\) (Exercise 8.1.14b).
## Computation of Some Simple Laplace Transforms
Find the Laplace transform of \\(f(t)=1\\).
###### Solution
From Equation \\ref{eq:8.1.2} with \\(f(t)=1\\),
\\[F(s)=\\int_{0}^{\\infty}e^{-st}\\,dt=\\lim_{T\\to\\infty}\\int_{0}^{T}e^{-st}\\, dt.\
onumber\\]
If \\(s\
e 0\\) then
\\[\\label{eq:8.1.3} \\int_{0}^{T}e^{-st}dt=-{1\\over s}e^{-st}\\Big|_{0}^{T}={1-e^{-sT}\\over s}.\\]
Therefore
\\[\\label{eq:8.1.4} F(s)=\\left\\{\\begin{array}{cl} \\displaystyle{1\\over s}, &s>0,\\\\ \\infty, &s0.\
onumber\\]
This result can be written in operator notation as
\\[{\\cal L}(1)={1\\over s},\\quad s>0,\
onumber\\]
or as the transform pair
\\[1\\leftrightarrow{1\\over s},\\quad s>0.\
onumber\\]
It is convenient to combine the steps of integrating from \\(0\\) to \\(T\\) and letting \\(T → ∞\\). Therefore, instead of writing Equation \\ref{eq:8.1.3} and Equation \\ref{eq:8.1.4} as separate steps we write
\\[F(s)=\\int_{0}^{\\infty}e^{-st}dt=-{e^{-st}\\over s}\\bigg|_{0}^{\\infty}=\\left\\{\\begin{array}{cl} \\displaystyle{1\\over s}, &s>0,\\\\ \\infty, &s0,\\\\ \\infty, &s0.\
onumber\\]
This result can also be written as
\\[{\\cal L}(t)={1\\over s^{2}},\\quad s>0,\
onumber\\]
or as the transform pair
\\[t\\leftrightarrow{1\\over s^{2}},\\quad s>0.\
onumber\\]
Find the Laplace transform of \\(f(t)=e^{at}\\), where \\(a\\) is a constant.
###### Solution
From Equation \\ref{eq:8.1.2} with \\(f(t)=e^{at}\\),
\\[F(s)=\\int_{0}^{\\infty}e^{-st}e^{at}\\,dt.\
onumber\\]
Combining the exponentials yields
\\[F(s)=\\int_{0}^{\\infty}e^{-(s-a)t}\\,dt.\
onumber\\]
However, we know from Example 4.1.1 that
\\[\\int_{0}^{\\infty}e^{-st}\\,dt={1\\over s},\\quad s>0.\
onumber\\]
Replacing \\(s\\) by \\(s-a\\) here shows that
\\[F(s)={1\\over s-a},\\quad s>a.\
onumber\\]
This result can be written as
\\[{\\cal L}(
