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The Laplace transform of a continuous function over the interval $[0, \infty)$ is defined by $F(s)=\int_{0}^{\infty} e^{-s x} f(x) d x$ (see the Student Project). This definition is used to solve some important initial-value problems in differential equations, as discussed later. The domain of $F$ is the set of all real numbers $s$ such that the improper integral converges. Find the Laplace transform $F$ of each of the following functions and give the domain of $F .$$$f(x)=\cos (2 x)$$

$F(s)=\frac{s}{4+s^2}$Domain of $F$ is $[0, \infty)$

Calculus 1 / AB

Calculus 2 / BC

Chapter 3

Techniques of Integration

Section 7

Improper Integrals

Integration

Integration Techniques

Missouri State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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the student project in your textbook defined the LaPlace transform over the region zero to infinity by this integral. So when we're given a function of eggs such as co sign of two X, we were able to plug in this function for f of X. So that will make it capital. F of s equals the integral from zero to infinity off exponential negative S X Times co sign of two X. So we need to keep in mind that we can't go all the way to infinity here. So we are going to have to use a limit so we'll have the limit as t approaches infinity of the integral from zero to t of exponential negative S X Times Co sign of two X with respect to X So now ignoring bounds. For now, we're going to look at just the integral of exponential negative s X tonnes co sign of two X with respect to X and then we can see what the bounds are later. So to solve this, we have to separate X terms being multiplied together, which is going to be difficult to integrate on their own so we can use integration by parts which you'll remember is the integral of you times DV mine equals u times v minus the integral of V times to you. So say you equal to exponential negative s ex and D V equal to co sign of two X. You could also reverse them. I'm just choosing to set you equal to exponential negative s eggs so that derivative is equal to negative s times. Exponential negative s ex and the anti derivative of co Sign of two X is equal to sign of two X divided by two. So now we're going to plug this in to our formula. So we have you ties V so 1/2 times one over Exponential s ex times Sign of two X minus the integral of v times Do you? So 1/2 time sign of two eggs, times negative s times, exponential, negative at sex. So we'll notice that we have two constants in our into girl. So I'm going to pull these out so that they don't get us confused here. So we're going to end up with one. We're gonna end up with sign of two eggs over two times Exponential s ex just simplifying some things on the outside plus s over two times the integral of sine of two x times Exponential negative s X, with respect to X. So we're going to use integration pipe parts again. So we'll set you equal to exponential Negative s ex again and Devi equal to sign of two x with respect to X so that the derivative of you to the negative SX is going to be negative s times Exponential, negative SX and the anti derivative of sign of two X is negative co sign of two X divided by two So now we complete this into our formula. So we have sign of two X over two times Exponential s ex plus s over two times u times v so negative 1/2 co sign of two x times one over exponential s ex minus s over to times the integral of you three times do you so 1/2 times co sign of two x times exponential negative s X and I ran out of room almost so again we'll notice that we have. These two isn't s here. We have these two constants on the inside of our integral. So we're gonna pull them out again so that we can have sign of two X over two times Exponential s ex plus rather minus s turns the coastline of two eggs over 24 times Exponential s ex combining things in that second term minus s squared over four times the integral of exponential negative S X Times co sign of two X So you'll notice that this is the same as theory Jinling Integral that we've that we had. So all of this is equal to the integral of exponential Negative SX Times Co sign of two X with respect to X. So I'm going to scroll down so that we can have more space to evaluate this. So if we add as squared fourths times this integral to both sides, we're going to end up with sign of two X over two times exponential s X minus as times co sign of two X over four exponential s ex equals one plus science s squared over four times the integral of exponential negative s ex Times co sign of two x with respect to X. So if we divide this term from both sides will end up with 1/1 plus as squared over four times. Signed of two x over two times Exponential SX minus sign s Times Co sign of two X over four times exponential SX as equal to this original Integral. So because our A rented original Integral had the bounds zero to t, we can now evaluate this from zero to t so plugging in tea and zero will have the limit as T is approaching Infinity of sign of two to t over two times Exponential of s T minus s Times Co sign of two t over four times Exponential ste times 1/1 plus X squared over four minus sign of two time 0/2 tens. Exponential s time zero minus s temps Co sign of two times 0/4 times Exponential s time zero times 1/1 plus as squared over four. So this is going to be equal to as we evaluate this as T is getting closer to infinity, this denominator of distraction gets larger and larger, bringing the whole fraction to converge to zero. So we'll have zero minus. Similarly, in the second fraction we have t is approaching infinity, causing the denominator to approach infinity, meaning the the fraction itself is approaching zero. So we have zero minus zero turns 1/1 plus as squared over four. So that's just gonna be zero minus sign of zero is equal to zero. So we have zero minus s times. Co sign of zero is over four times. Exponential zero is going to be s times 1/4 times exponential zero, which is also equal toe one. So s times 1/4 times, one times 1/1 plus as squared over four. So if we multiply these together, this is gonna be positive. Will have s over four plus s squared as our solution, So f of s is going to be equal to s over four plus as squared. And now we need to find domain for us. So one thing to notice is that we do have in s in this in the denominator of our fraction, however, we're adding a square of that s so over our original domain of zero to infinity, there will never be an s for which this fraction is not does not exist

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