Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. The Laplace transform Given a function F of T defined for all T greater than or equal to 0, the Laplace transform of F is the function F defined as capital F of S equals the Laplace of the function little F of T which equals this integral of 0 to Infinity of E to the negative St. times the smaller function FTDT for all values of S for which the improper integral converges. If we thought about letting F of T equal 1, we'd have the Laplace of 1 equaling 0 to Infinity of E to the negative St. times one DT. Now that turns into the limit as B approaches Infinity of the integral zero to B of E to the negative STDT. And we know that that equals the limit as B goes to Infinity of -1 / s E to the negative St. from zero to B. So when we stick in our bounds, we get limit as B goes to Infinity of -1 / s E to the negative SB plus 1 / s E to the negative s * 0. Now E to the zero at the end is really just one. So what we need to think about is this first piece. And if we think about B is going to Infinity negative S times Infinity. So E to the negative Infinity would be one over Infinity E to the Infinity. And that would get really, really small. So it's contingent on South being positive. So this whole first term is going to go to 0 if S is positive when B goes to Infinity. So then we'd really just end up with 1 / s for S greater than 0. It's going to diverge or not exist if S is less than or equal to 0. If you thought about S being a negative number, we'd get a negative times a negative which would be positive times Infinity. So E to Infinity is something really really big and hence that would diverge. If we look at F of T equal E to the AT where T is greater than or equal to 0, the Laplace says E to the AT. So by our definition, limit as B goes to Infinity of 0 to BE to the negative Ste. to the AT DT. When we have the same basis, we're going to add the exponents and I'm going to factor out a negative T. And the reason I want to factor out a negative T is when I stick in Infinity. We know E to the negative Infinity is going to be something really, really small and basically go to 0. So we're going to get with some manipulation limit as B goes to Infinity of -1 / s minus AE to the negative TS minus a from zero to B. So that first term is going to go to 0, and we're going to end up with just 1 / s -, a when s -, a is greater than 0. And I need that to be greater than 0 because this exponent back here needed to be positive so that I'd have negative Infinity times some positive number to make the E to that power all go to 0. So the Laplace says E to the AT is equal to 1 / s -. A when S is greater than a gamma function is going to be defined as equaling 0 to the integral zero to Infinity of E to the negative TTX -1 DX where X is greater than 0 where gamma of one is equal to 1 and the gamma of X + 1 is going to equal X times gamma of X. So if we thought about the gamma of n + 1, we could realize that that equals N times gamma of N, which would then equal n * n - 1 of gamma n - 1, which would then equal n * n - 1 N -2 times gamma n - 2. If we continue that with this, we can see eventually that this is really just going to be N factorial. So now if we're going to do Laplace transform of a power function F of T equal T to the a, we get our limit as B goes to Infinity of 0 to BE to the negative STT to the ADT. We're going to do AU sub here. We're going to let U equal St. so that DU would equal SDT or one over SDU is DT. And if U equals St., we know that U / s equal T So now when we do our substitution, we're going to get limit as some new variable C goes to Infinity of the bound zero to C I'm sticking in zero to B into that substitution. So then I get E to the negative U times the quantity U / s to the a * 1 over SDU. The S to the a * s is just going to be S to the a + 1 of the denominator. And S is a constant because it's not being the derivative or the integral in terms of U. So I'm going to be able to pull it out of the integral sign. So the limit of C goes to Infinity of 1 / s to the a + 1 integral zero to CE to the negative U * U to the Adu. Now this was really just our gamma function from a minute ago. So this ending piece is our gamma function. So we're going to get 1 / s to the a + 1 gamma of a + 1, and the gamma of a + 1 is just a factorial. Recall from back here that our gamma was the function 0 to Infinity integral E to the negative TT to the X -, 1 DX where X was greater than 0. So the gamma of n + 1 was just N factorial. So when we get here, the gamma of a + 1 is just a factorial over S to the a + 1. Once again, this is only true if our S is greater than 0, because I needed this T back here when it was going to Infinity for the E to the negative St. to go to 0. So a given value for gamma of 1/2 is going to be sqrt π. That's just a given. So if I wanted the gamma of three halves, that would be 1/2 gamma of 1/2. But gamma of 1/2 is sqrt π, so we'd get 1/2 sqrt π. So now we can evaluate things that have fractions as far as the gamma function goes. So hyperbolic cosine of KT. The definition of hyperbolic cosine is E to the KT plus E to the negative KT over 2 where K is greater than 0. If this had just been some variable a, it would have been AE to the a + e to the negative a / 2 where a is greater than 0. So if we want the Laplace of the hyperbolic cosine of KT as limit as B goes to Infinity, we go zero to BE to the negative St. E to the KT plus E to the negative KT over 2 DT, I'm going to factor out the half. We're going to have zero to B. When the exponents are the same, we add the exponent. When the bases are the same and we're multiplying, we add the exponents and I'm going to factor out the negative TS. So then when I take the integral and simplify, we're going to see that we end up with net limit as B goes to Infinity of 1/2 negative 1S minus KE to the negative TS minus K -, 1 / s plus KE to the negative TS plus K from zero to B. So when I stick in my upper bound of B, this negative B is going to go to negative Infinity and E to the negative Infinity is going to be really, really small going to 0. That's assuming that this s -, K is positive and also that this s + K is positive. So this first term is both going to go to 0 these first terms. So we end up with 1/2 of 1 / s -, K + 1 / s + K If we get a common denominator here, we'd get s + K + s - K over the denominators multiplied together. Simplifying we get s / s ^2 -, K ^2. OK, so before the phone ring we were getting common denominator and simplifying. So the two and a half are going to cancel. We get s / s ^2 -, K ^2 where the s -, K is greater than 0 and the s + K has to be greater than 0 because I need both these exponents to be positive. Now in the original we were told that K was greater than 0. So if we solve these we can see that S has to be greater than K which had to be greater than 0. We know that the hyperbolic sine of KT is E to the KT minus E to the negative KT over 2, cosine of KT in terms of ES is E to the ikt plus E to the negative ikt over 2, and sine of KT is E to the negative KT minus E to the negative ikt over 2 I. Now just a refresher, we can remember that if we graph in terms of the real and imaginary planes, we'd have X going over the real number and Y going up the imaginary number and R being our distance out to some point XY in the real imaginary plane. So Z would just be a point. Or actually it could be considered a vector even of X plus IY where we went over X and we went up IY. So if we look at this, we get X equal R cosine Theta and Y equal R sine Theta. So R-squared is X ^2 + y ^2. Or we could think of our Z as R cosine Theta plus IR sine Theta. If we factor out an R, we get Z equaling R cosine Theta plus I sine Theta. Recall that Euler's formula says E to the I Theta is cosine Theta plus I sine Theta, which is what we see right here. So we could think of that Z as RE to the I Theta. But we're going to use this definition of E to the I Theta equal cosine Theta plus I sine Theta. Thank you and have a wonderful day.