 So, this is where we were last time and we want to prove this theorem ok. So, let us see how do we do this. What I will do is look at that line which I just defined and so, what I am going to do is to take define two domains they are both rectangles they are the common part of this rectangle is this vertical line at C and going up to plus r up there minus r down there and this extending on the this particular domain is a rectangle extending up to plus u here and this domain d 1 is going up to minus u ok. And the reason I am defining two domains is because I want to show a certain behavior of this function which is very different depending on the value of x. So, below one it is one behavior above one is another completely different behavior. So, let us see. So, if you recall the integral that we are looking for is x to the s by s d s right. So, let us first integrate it over or around the boundary of d 1. What is this? This should be easy the integral 0 y in d 1 it is not analytic at s equals 0 it has a pole at s equals 0. In fact, this is the only pole for this right there is no other pole. So, inside and 0 is inside d 1. So, that is a pole at d 1. What is the residue at this pole? You multiply this function x to the s divide by s by s and take the limit s s goes to 0 and residue is 1. So, therefore, this is this integral is 2 pi i fine. So, that is what we know directly from the residue theorem. Now, we can now split this integral as before into because over a rectangle. So, we will split it into 4 parts the first part goes from c minus i r to c plus i r the second part goes from minus u plus i r c plus i r. So, I am traversing the domain in counter clockwise version right c plus i r to minus u plus i r plus minus u plus i r to minus u minus i r and finally, minus u minus i r to let us call this i 2 i 3 and the strategy will be same as before this is the integral of interest that we want to show something for and so, we want to get rid of the remaining ones. Let us try to do that and the same apply the same strategy. Let us look at the absolute value of these integrals. So, what is the absolute value of i 2? That is the absolute value of this integral c plus i r to minus u plus i r x to the s by s d s this is less than equal to integral of absolute value. What is absolute value of s? As s goes from c plus i r to minus u plus i r. So, s is going from here to here what is the absolute value of s? It varies surely, but it is at least r and that is all we need because we are dividing by absolute value of s is at least r. So, this is at least and what about x to the s? x to the s is e to the s log x d s c plus i r to minus u plus i r and note here that although s is a complex number the only part that varies in x s is the real one. So, I can rewrite the same integral as by the way this is taking the absolute value. So, s I can write as real part plus the complex part when you take the absolute value the complex part anyway goes away because that is absolute value is 1. So, only thing I left with here is 1 over r c to minus u e to the t log x and there should be somewhere outside I should just put absolute value I do not want any negative values anyway. And this can be integrated easily what is the integral of this e to the t log x d t e to the t log x divided by log x going from c to minus u and this is less than equal to well the bigger part would be when t equals c that is x to the c by log x minus x to the minus u by log x. And by let us put a plus also it does not matter just to be on the same side. So, that is the estimation for I do and if you look at this one I 4 is the same thing it was the integral going up and this is the same along the same line just reflecting down in the imaginary side along the real axis. So, it is the same the absolute value wise it would be the same. So, you can write bound this also that leaves I 3. So, let us do the same exercise for I 3 that is from minus u plus I r to minus u minus I r what is the absolute value of s. So, here it is at least u. So, we can just use that at least u the absolute value d s and now again s breaks into real and imaginary part when you take the absolute value the imaginary part vanishes what about the real part how big is the real part is exactly minus u. And so this is less than equal to 1 by u minus u plus I r to this becomes x to the minus u. So, there is really no integral to be done and this is a trivial integral and therefore and this integral is at most of lengths 2 r. So, in fact when you take the absolute value that becomes at most 2 r. So, that you get basically 2 r x to the minus u. So, plus this 2 times this plus this. So, that is equal to I should stick an order here plus x to the minus your log x. Now, let us take the limits and the first thing we do is we send u to infinity this is of course, 2 pi i. And as you send u to infinity what do you get because u does not occur in in the here and anyway. So, I can easily send u to infinity what comes or what remains in the error term 2 of course, I can hide in the constant this goes to infinity this goes to infinity. And now if I send r to I should be careful. So, when I send u to infinity what happens this will happen only if x was. So, this is a very important clause to be remembered as long as x is greater than 1 this is what happens if x is was less than 1 then this will go to infinity that will be pointless. So, assuming x is greater than 1 I get this approximation and now send r to infinity if x is greater than 1. So, that is the result of integrating on this domain. Now, let us look at the result of integrating on d 2 it will be pretty much similar first of all by residue theorem what is the residue 0 is completely analytic on this domain. So, this is 0 you see that is the reason we chose these 2 domains here because of the pole being here we get a residue which is 2 pi i or divide by 2 pi i get 1 here is completely analytic. So, it goes to 0, but also the important thing is it to sort of coincide this 0 1 with the value of x what I did was to look at this specific integral. So, this says that the integral thus converge, but only when x is greater than 1. Now, let us look at these this one and same way we will express this as sum of 4 integrals again we let this to be i 2 i 3 and i 4 i 2 and i 4 are again similar integrals in absolute value. So, their absolute value is going to be less than again using the same ideas x to the s is e to the s log x and now absolute value for s is going to be how much at least r is going from far away right the absolute value here. Again the imaginary part vanishes and then you can this we can convert to a real integral which goes from c to u e to the t log x 1 by r comes out here dt and this is less than equal to x to the u by log x plus x to the c by log x right and same with actually i 4 absolute value i 4 also as the same property and what about i 3 i 3 is u plus i r to u minus i r well s is going to be absolute value of s is at least u and absolute value of x to the s is going to be x bar u s and this again is 2 by 2 r by u now send u to infinity what happens what happens to the first term well depending on x if x is less than 1 this goes to 0 if x is bigger than 1 this diverges. So, we focus on x less than 1 and we get that this is equal to same thing with the second term in fact this is the absolute value of x. And now if you send r to infinity we get what we get this is 0 there is something missing this is 1 by r on the outside for i 2 and i 4. So, this also goes to infinity this gives me therefore the function that slightly changed because this is when x is between 0 and 1 this is 0 and then steps up and then stays at 1 forever what we wanted was 1 between 0 and 1 then stepped on but that is trivial to change slip. So, it plays x by 1 by x and you get the other function. So, this is going to be it is very interesting function to begin with and it is going to be very useful for us later on also it demonstrates the power of this contour integrals that you can use the knowledge about the poles inside a contour actually evaluate the integral very easily at x equal to 1 here the x equal to 1 this what let us see what happens if x equal to 1 at x equal to 1 you get d s by s right d s by s is log s log s integrated from c minus i r to c plus i r what is that that is firstly it we have to think about what what is the interpretation of log s we are taking we have already discussed it a lot the nice thing about here is that because the line is on the positive side it is not cutting any of those funny things on the left hand side. So, it is only on the one single sheet and it is going from c minus i r c plus i r and so we can just use the prints the main value of our log s which is so log of c plus i r minus log of c minus i r and then that becomes log of c plus i r divide by c minus i r and as you send r to infinity what do you get you get i divide by minus i and minus 1. So, you get log of minus 1 what is log of minus 1 log of minus 1 is i pi that does not sound right no that is right that is about right. So, this is let me just say exercise that and therefore, 1 over 2 pi i c minus i infinity to c plus i infinity x to the s by s ds this equals 0 if x is between 0 and 1 half if x equals 1 1. So, this is the step function for us any questions good. So, I am pretty much done with introduction to complex analysis now I would like to start on the real content of the course open this up 15 minutes. So, I would like to start it from the next class. So, to fill up the time let us do one more integral this time let us do something different now who can integrate this do you know this integral classically how do you solve define a sin alpha that will require a to be between minus 1 to plus 1 1 upon and then do something with it then you can sin alpha suppose is sin alpha plus sin theta. So, that is one way of doing it somewhat messy. So, here is a much simpler way of doing it same strategy think of this integral as happening over complex. So, first I want to translate everything over the complex numbers. So, here theta is varying 0 to 2 pi let us imagine that we are there is we are integrating over complex same theta 0 to 2 pi. But now instead of theta we can actually view it as a all the points we are integrating over our complex numbers. So, let us write this as first let us do the substitution. So, let z be e to the i theta which is therefore cos theta plus i sin theta. So, what is sin theta in terms of z what simpler z bar no you do not want to get z bar z bar is by the way did I talk about this that the function z f z equals z bar is it analytic or not we have an is it analytic f z equals z bar how do you prove it to be analytic Cauchy Riemann do Cauchy Riemann. So, continuities are assumed that is simple does it satisfy Cauchy Riemann it does let us pull out an intermediate pitch or I can write f of x plus i y equals. So, what is Cauchy Riemann this is u this is minus y is v. So, del u by del x is del v by del y minus fails del u by del x must be equal to del v by del y. So, not analytic. So, this function is not analytic. So, which means that we cannot use z bar anywhere in analytic function essentially. So, coming back to this z bar is no go, but that idea is good that you want to get cos theta minus i sin theta somewhere. So, that you can subtract and get just sin theta how do you get cos theta minus sin theta sin minus i sin theta without using z bar e to the power minus i theta exactly. So, which is e to the minus i theta is cos theta minus i sin theta which is very good, but what is e to the minus i theta in terms of z is 1 by z it is very fine. So, now you subtract and so we get sin theta is therefore, 1 over 2 i z minus 1 by z we are choosing that we are choosing z to be e to the i theta making sure that mod z is 1. Because we are integrating over the circle and we can choose any radius of the circle one radius one is the best for us good. So, that sin theta and what is d z is i e to the i theta d theta which means i z. So, now we got everything to do the substitution and I can write d theta is d z by i z and sin theta is 1 over 2 i z minus 1 over z and this is equal to d z over a i z plus 1 over 2 z square minus. So, that is what the integral looks like and this is over a circle of radius. So, what is the value of this integral or it depends on whether there is a pole inside the unit disc or not. So, and that is easy to determine it is a quadratic in z what are the two roots of z they are minus 2 a i plus minus square root of minus 4 a square plus 4 right divide by 2 and therefore, you get minus a i plus minus. So, where are these roots first of all it is real and imaginary part are very clear real is either plus or minus square root of 1 minus a square and imaginary part is minus a i it depends on a yes absolutely it does depend on a. So, what happens if a is greater than 1 then there is no pole inside and then the integral would be 0 and what happens if the a is between minus 1 and plus 1 then it will have will have both the roots in yeah it will have both the roots inside wait is that necessary this square plus this square is 1 a greater than 1 this probably is not even does not even make sense when this this process will fail if actually a is bigger than 1 because then if he is greater than 1 then this actually becomes complex which is just makes things messier than before no that is then it is not necessary that will be outside if a is bigger than 1 then this would be i times something i times a real number. So, then you get either add or subtract that real number to a so you might get inside or outside the circle. In fact, if a is large one of them will certainly be inside the circle because that real number would be very close to a square root of a square minus 1 if a is large that is going to be very close to a. So, when you both get added that falls out if they once get subtracted from the other that would be inside and if a is between minus 1 and plus 1 then this is both are inside. So, we split this in therefore, 2. So, let us analyze the first if minus 1 is less than a is less than 1 then what happens. So, then the integral would be what let us write this instead of first doing it let us factor this is a factorization and therefore, this is equal to at this the residue is one is. So, basically eliminate this and substitute for z minus a i minus square root of a square which will be 1 by 2 minus 2 square root of 1 minus a square plus when you substitute this what is going to be the residue is this correct and this is 0 this cancels with this and if a is bigger than 1 then going back one of them is the residue. So, when this is plus that is the only residue that survives. So, the one that survives is this one that is the pole inside and though you make this substitution and then you get minus which is of course, wrong why is this wrong it should be a real number this is complex 1 minus a square is complex what excellent. So, this is basically 2 pi by square root of a square minus shouldn't there be a minus now there is a minus already here and this when you take this or there should minus 1 square root is i which gets cancelled. So, there should be a minus or it could have been a plus 1 or minus 1 depending on whether square root of minus 1 you choose to be plus i or minus i probably is minus 1 directly this this one has to look at and see what is the right value may be going back to the original integral a is greater than 1 that is right and sin theta varies between minus 1 and 1. So, everything is positive. So, this must be positive. So, there you are that is the solution of the integral then in a very straight forward way no clever substitution needed just apply the blindly the residue theorem. And not even of course, this is just now that we are playing around let me just state another sort of obvious thing also that if this a was a complex number instead of a real number greater than 1 a complex number with absolute value greater than 1 then what would happen. So, this is an analytic function of a right it does not have there is no pole in because the absolute value of a is greater than 1. So, there is no pole of this and we saw that one of the theorem we proved was that in if there is an analytic function its integral is also analytic right. So, this is an analytic function in a which on the real line positive real line a greater than 1 agrees with this therefore, it must agree with it over the entire complex that half plane absolute value a greater than 1. So, you get that result for free of course, this is also analytic for absolute value a greater than 1 both are analytic both agree on line and therefore, they agree everywhere. So, that is it for today.