Feynman's Technique of Integration

first of all let me tell you i got this question from this book right by dailymath as i told you guys last time so check that out if you would like i will have the link in the description for your convenience all right i'm going to start by calling this to be f and let's call the parameter alpha and this right here we'll just write it as integral from zero to infinity e to the negative x squared cosine but instead of the five we are going to put down alpha and then x dx because look at this f of 5 is exactly our original integral and in fact after we finish the procedure right here you will see you can compute this kind of integral with whatever number you want right here very nice okay now here is the time to do our derivative on the left-hand side and also the right-hand side we are going to differentiate both sides with respect to alpha so let's put on d d alpha on both sides on the left hand side i'm just going to write down the derivative in the prime notation so here is f prime of alpha however on the right hand side we will have to bring the derivative inside of the integral and that's called the finance technique or you can also call that wbd differentiation under the integral sign when you bring this in because inside here we have both the alpha and also the x we have two things right so be sure you change that to partial derivative and let me just make the alpha slightly more clear for you guys all right so here is the integral from zero to infinity and then again we are looking at this as partial with respect to alpha and then all that so we have e to the negative x squared cosine of alpha x and then of course on outside we still have the dx like this now we are going to work this inside out so have a look right here integral from 0 to infinity in the alpha world x is a constant so is e to the negative x squared of course we will just write that down first here is the e to the negative x squared now as we can see alpha is inside of cosine so we are going to differentiate cosine and that will give us negative sine and the input stays the same but channel says we have to multiply by the derivative of the inside with respect to alpha alpha is the variable x is the constant so we have to multiply by x right here very nice and of course this right here we still have the dx on the very outside like this all right now this is so wonderful because we can integrate this part and we can of course differentiate this part so let's go ahead and do the di setup for the integration by parts i'll put it down right here d and also the i plus minus to get ready integrating this part negative x e to the negative x square differentiating sine of alpha x keep in mind though we have to do the work with respect to x right when the x world now so differentiating sine alpha x we get cosine of alpha x and then the general says multiplied by the derivative of alpha x with respect to x this time we multiply by alpha and then for this one we are just going to do the integral in our head we end up with one half e to the negative x squared thanks to this x right here on the outside we can just use up for that very nice now to continue of course we know this times that it's the first part of the answer so i'm going to just come back right here and we're going to write down well i'm going to put this on the top so we have sine of alpha x and then over we have the 2 and also bring that down because of the negative exponent so we have 2 e to the positive x squared that's the first part of the answer so don't forget to plug in plugging right from zero to infinity and then to continue remember we have to multiply this row and this right here is still an integral negative alpha and then we have the one half so let me just write down minus alpha over two the constant in the front and then put this and that itself the integral right so we have the integral still going from zero to infinity and we have this first because why not and you'll see we have a small draft full because doesn't this look familiar yes it does it's our f of alpha alright so now look at this this right here is super nice because the answer to this is zero when you put infinity to here like e to the infinity and then you square that man the button is so much bigger than the top the top is at most one right one negative one like yep well you get zero and then when you put zero into here thanks to sum of zero you get zero as well so we end up with this this right here we have zero and then i will put down minus alpha over two and this right here is our f of alpha so we will of course just write that down and now remember the left-hand side we have f prime of alpha and then on the right-hand side we have negative fish over two times f of alpha like this right yes this right here is a differential equation but don't worry this right here is a separable differential equation so it's not so bad i'm going to just do the work right over there for you guys i'm going to write this as d f d alpha right so df over d alpha and this right here is equal to negative alpha over two and then multiplied by f like this right and then separate the variable divide the f on both sides multiply the df of d alpha on both sides so we have d f over f this is equal to negative alpha over two times d alpha and then of course we can integrate both sides now very nice the left hand side we get ln absolute value of f plus a constant but don't worry just do it on the right hand side integrating this add one here which you get two and then divided by two so we end up with negative alpha squared over four plus our first constant now to isolate f let's do a lot of things in your head do e to the power on both sides e to this power is the same as saying e to the power times e to the c 1 but c 1 is a constant e is a constant c1 e just constant e to the c1 is the constant very nice and then do the absolute value you have to take it out of the absolute value to the plus minus so you have another constant anyway i'll just write this as f that's equal to our constant i'll just put it as c2 and then we have e to the negative alpha squared over four like this right well this is actually our f of alpha and we solve this differential equation this is very nice but the problem is that we have a c2 right here oh well this right here refer back to our original integral and hopefully we know something about that so have a look this right here is the same as say integral from 0 to infinity e to the negative x squared cosine of alpha x dx and this is equal to c2 e to the negative alpha squared over 4 like this well what do we know about this the truth is we never like the cosine of alpha x in the first place right so now let's think about how can we get rid of this oh well in this situation let's go ahead and let alpha to be zero because if we put zero inside cosine of zero is just going to be one very nice and of course we can also put a 0 right here so have a look this is going to give us the integral going from 0 to infinity e to the negative x squared only because this part will be 1 when alpha is equal to 0. on the right hand side we just get c2 because when you put zero right here e to the zero power is one so that's very nice and now what's this the gaussian integral this right here is just half of that right so this right here is famously equal to square root of pi over 2 which is the c2 right there so now this right here square root of pi over 2 we can just put that back and we get our f of alpha so i will go back here for you guys i will tell you our f of alpha which is equal to our original integral right here in terms of alpha this in fact is equal to square root of pi over 2 and then times e to the negative alpha squared over four and now depends on what alpha value that you want you can actually end up with a lot of very nice looking integral our original one of course is when alpha is equal to 5. this right here is nicely equal to square root of pi over 2 e to the negative put the 5 in here so we have 5 squared over 4 of course this is square root of pi over 2 e to the negative 25 over 4 very very nice you see you have the e you have the

Channel: blackpenredpen

Views: 528,945

Keywords: Feynman's Technique of integration, differentiation under the integral sign, integral of e^(-x^2)*cos(5x) from 0 to inf, feynman's trick, math for fun, blackpenredpen, interesting integral

Id: YO38MCdj-GM

Length: 10min 2sec (602 seconds)

Published: Sun Dec 15 2019