Integration By Differentiating Under The Integral Sign (HBD Feynman)

what's going on smart people now I actually already just recorded a video for today but I just found out that it's Richard Feynman's birthday so needless to say that video is going out tomorrow and today is all about Richard Fineman more specifically one of his favorite methods of integration differentiation under the integral sign so the way we're going to start this off is I'm going to take an integral that I found online that withstands all of the tests and procedures of usual integration methods like by parts and the use substitution and then we're going to throw differentiation under the integral sign at it and we're going to come out with an answer so the integral that we're gonna be solving today is the integral from 0 to 1 of T cubed minus 1 over Ln of T with respect to T now right off the bat you might notice that the thing that makes this integral so ugly is this factor of Ln of teen the denominator and whenever you just have one thing one little thing would be pretty nice if that wasn't there that's when you might as well give differentiation under the integral sign a chance and the way we're gonna go about doing this is by parameterizing a part of this function and making it a function of something other than T and then what we're gonna do is we're gonna take the derivative of that with respect to that other variable and see if that's something we can exploit in this actual integral let me make some context here so let's say we take a look at some function T to the X and we want to take the derivative of this function that depends on X with respect to X T to the X ok well that's just gonna equal to the X still it's just an exponential times Ln of T now if t T here we're pretending it's just a number right this could be e and then if this was a then it would just be the derivative of e to the X so that would just be e to the X still times Ln of e which would give us just e to the X that works cool now with this in mind what we're going to do is we're going to define a new function as an integral so we're gonna define some function G of X that is equal to the integral of T to the X minus 1 over of T DT so you can see and we're gonna have this going from zero to one and if we integrate this while we're adding up all the things that depend on T the only thing that's left over is something in terms of X that makes it a function of X okay now what we're gonna do is well first off let's look at this case let's look at the case that we look at G of three so if we have G of three then what we get is that that X becomes a three and we just get this integral here so really I'm just defining this in a bit more abstract in general way what I want to find now is I want to find what G prime is G prime of X and now this is the function of X this means that I'm taking the derivative of this with respect to X okay so what we're gonna do it if we haven't added up all the things that depend on T yet that means when we take the the total derivative with respect to X and put it inside the integral sign it becomes a partial derivative so I want to take the derivative of this so it's going to be the partial derivative with respect to X of T to the X minus one over Ln of T with respect to T okay well we could split up this part here into two fractions or into two parts right it could be to the x over ln t minus one over ln t and when we attack it with this derivative with respect to x maybe it's best that i read it out that's the same thing as the derivative is this d DX minus d DX DT i just split up this fraction and what we see is that this part goes to zero right because this doesn't depend on X at all that goes to zero and we just need to find out what the derivative with respect to this is well we just solved that up here so this part comes out two that tells us the crime is equal to the integral from zero to one the derivative with respect to X of T to the X is T to the X Ln T so we get T to the x over 1 / Ln T DT the l ends cancel and we just get that this is equal to the integral from 0 to 1 of T to the X DT okay well this is just you know power rule that's nothing that we don't know how to take care of so that's just going to come out to T to the X plus 1 over X plus 1 and we're evaluating this from 0 to 1 and what that comes out to well remember that this 1 and this is zero is corresponding to these values of T so that's just 1 to the X plus 1 over X plus 1 minus 0 to the X plus 1 but that's just 0 so I'm not going to need to write that not to mention want anything to the any power 1 to any power is just going to give us 1 that tells us that this is equal to X plus it's just 1 over X plus 1 and that is equal to G prime of X ok now I'm going to erase some of this and we're going to see how to connect this to the initial integral that we wanted to solve so now we have a actual expression for G prime of X and what we're going to do now is we're just going to work backwards we're going to take the integral of this with respect to X so that tells us that G of X is equal to the integral of I guess the antiderivative of 1 over X plus 1 with respect to X this isn't a crazy integral you should recognize this as just being a line of X plus 1 and we're doing an antiderivative here so this is going to have a factor of C okay cool so this is saying that these two expressions are equivalent this integral of this ugly thing here is coming out to this plus some factor of C and we need to find out what the C is and we can find out what the C is just by looking at this function here and noticing that if we put in G of zero so if we let G of zero what is that let's see what G of zero is well we get anything something to the zeroth power which is just going to be one so we get the integral from zero to one of zero times DT which is zero so we have an initial condition that we can use to fix this factor C here so what we can do is we can plug in zero for X here and solve for what C must be so if G of zero equals zero the G of zero is equal to Ln of one plus C that equals zero well Ln of 1 is just zero which tells us that C equals zero okay so that tells us that G of X is equal to Ln of X plus one and the case that we were interested here in here is the case where it's G of three right because that's the initial integral that we wanted to solve because if X is three then we get this instead of this general expression here so that tells us that G of 3 is just this evaluated at 3 which is just going to be Ln of 4 great so once it's all said and done that tells us that this integral that we were struggling saw in the first is equal to Ln of 4 thanks Fineman now what made me make this video was two reasons one is that someone in the comment section reminded me that this fineman's birthday and someone also commented that an integration technique that they would like to see his differentiation under the integral science so thanks to you two for putting this idea in my head in the first place hope you guys found this method interesting or a little bit helpful let me know in the comment section if you did and I'll see you guys there

Channel: Andrew Dotson

Views: 251,104

Keywords: andrew, dotson, physics, integration, calculus, richard, feynman, technique, integral, daily upload, 58, daily physics upload, odu, leibniz, happy birthday, big bang theory

Id: pqP13eLG35U

Length: 9min 4sec (544 seconds)

Published: Fri May 11 2018