Feynman’s Integration Technique is Overpowered…

hello everyone today we're going to be integrating s of X overx DX evaluated from zero to Infinity using internet's uh most favorite integration technique I must say Fan's technique okay now it has a few names F technique or differentiation under the integral sign is it it has a few names um referenced in pop culture for example like uh The Big Bang Theory um anyway so um yeah so we're going to be integrating s of X overx evaluated from0 to infinity and of course there is the beautiful man himself Richard fan down there okay um I have actually recorded this video already and I finished it and realized I did not record my audio so that was brilliant anyway um I have checked and the record the audio is recording this time anyway so we are going to be looking at the integral from 0 to Infinity of sin of X /x DX now I I urge you to to attempt this using the sort of um standard integration techniques that you may learn at a level maths or a level further maths and honestly best of luck to you it is um don't know if it's possible or not to be honest um anyway um go ahead and and and try um we're going to be using Fan's technique anyway and Fan's technique involves defining an integration function uh with a new sort of PR parameter or or or a new variable if you will okay and that function that we are going to Define I of a is going to be the integral from 0 to Infinity of sin of x /x * e to the a x like so and what I'm going to do actually is change those letters uh those letter A's to the yellow color so you can see them a little bit better there we go um and the reason we are do well first of all that negative there in front of the ax that will become apparent later on as as for the reason we have that there um but as you can see um we are concerned about when I is evaluated at zero when when a is equal to Z because if a is equal to zero the exponent here becomes zero e to the power Z of course becomes one and so I to the uh I evaluated what am I talking about y of 0 is equal to the integral from 0 to Infinity of sin x/x DX like so um so that's what we are going to go ahead and do um so first things first well we've defined our function let's go ahead and differentiate both sides um with respect to a so I of a differentiated with respect to a is of course I dash of yellow color a okay and well what happens when we differentiate the right hand side with respect to a well we are attempting to uh differentiate through an integral almost when our variable is trapped inside of the integral so of course we are going to be using partial derivatives so we're going to sort of bring the derivative inside if you will and uh differentiate partially with with respect to a so this is going to be the integral from 0 to Infinity of the partial derivative with respect to a of sin of x/x * e a x DX um oh yellow color for the a of course my bad I did forget that there is a and a okay so well we are differentiating partially with respect 2 a so everything that isn't an a is just a constant so sin of x/x in the a world is just a constant and this x right here in the a world is just a constant and so the derivative of this function is actually going to be fairly simple it's going to be the integral from 0 to Infinity of sin of x/x stays as it is it's just a constant times the negative time x e to the a x okay and this arises because of course if you take the derivative with respect to X of e to the KX where K is a constant we get k e to the KX similarly if it's a negative okay that netive K just comes down to the front that's exactly what's happened here we've got this negative X that has just found its way down at the front there okay so as you can see that X and that X cancel very very nicely and so we get I dash of color a is equal to the integral from 0 to Infinity of s of X oh it doesn't need to be up there anymore sine of x I can write neater than that looks like a spider has died on my on my iPad s of x e to the a x DX of course okay well this integral is significantly easier for us to solve okay because we can just do this by parts so that's exactly what we are going to do I've also just forgotten that we have a negative sign there that negative sign is going to make its way to the front of this integral okay I do apologize for forgetting that um so please don't write angry comments in the comment section for missing for me missing that's when out um I have corrected it anyway so yeah significantly easier for us to solve now let's go ahead and grab this integral and bring it over here I'm going to find uh Define J to be the integral from 0 to Infinity we'll deal with the negative a bit later on uh the integral from 0 to Infinity of s of x e to the ax and DX okay now we're just going to do uh integration by part so U is equal to sin of x v- equal to e a x u- is equal to COS of x v is equal to 1 1 a e a x okay and using the integration by parts formula UV minus the integral of V du we of course get um -1 / a sin of x e a x um minus the integral of V which is 1 a e to ax so that negative can come to the front to join with this negative here and create a plus and that one over a can can can join in the front as well e ax and of course we Times by the derivative of U which was COS of X so there we go well the we have we got any closer yes we have CU we can just do integ ation by Parts again we can say our COS of X is U and our dvdx is e to ax okay so U is equal to cosine of x v- is equal to e the ax so therefore the derivative U is netive s of X and V is -1 / a e a x like so and then of course just uh employing the d uh integration by parts formula again we get -1 / a COS of x uh e a x um plus no it's not it's minus the integral of 1 / a sin of x e to a x DX which should alarm Bell should be ringing here because this right here this the integral of s of x e ne e toga ax is just what our J was so we can write that in this is equal to J or1 over J for that matter okay so let's write in what we have so far so we have got J is equal to we're just going to bring this down -1 a s of X sin of x e to the a x + 1/ a Time Time -1 / a cine of x e a x - 1 / a * J where J was our integral okay then so expanding this out we get J is equal to -1 a sin of x e to a x - 1 / a 2 cosine of x e to a x - 1 / a^ 2 J let's go ahead and times up by a 2 on both sides we get a^ 2 J is = a sin of x e to a xus cine of x e a x - J let's go ahead and add J to both sides so we get a^2 + J is equal to all of this so a sin of x e to a x - cosine of x e to a x well we can oh sorry that should be a j there my bad um let's factor out J now so we get J * a^ 2 + 1 is equal to um let's factor out netive e to the uh negative ax so let me bring this down actually a minute down you go right negative e to the ax and then what well what we going to have inside of those brackets well we've got a sin of X Plus cine of X let's divide by a 2 + 1 now to get J J is equal to e to ax a sin of x + cine of X over a 2 + 1 okay fantastic now J is equal to this and if you recall from earlier well why did we create J in the first place well it was to evaluate this integral right here this integral had a negative on the front so we know that I dash of a okay is equal to negative J so netive J where J was this right here okay so I of a is equal to um e to a x a sin of x + cine of X over a^ 2 + 1 now of course um J was was the integral from 0 to Infinity right there it was so let's go ahead and evaluate those limits okay so from 0o to Infinity so i- of a is going to be the limit and let's choose a letter let's choose the letter um t Okay t uh so the limit as T approaches Infinity of e to a x a sin of x + cine of X over a 2 + 1 evaluated from T from 0 to T okay well let's have a think what happens when uh we substitute T in and T approaches Infinity okay well when we substitute T and we're going to get e to a t like this and then dot sort of the a sin x cosine X over a^ 2 + 1 okay but more importantly let's think about what happens as T approaches Infinity okay so as t approaches Infinity e to the a t which can be written as 1/ e to the a t well as T approaches Infinity this approaches infinity and one over and one over um X as X approaches Infinity for example approaches zero so our um e to the netive a as T approaches Infinity approaches zero which means that this entire thing approaches zero when we substitute T into it okay because this e to the negative uh ax is going to uh sort of decrease faster than a sin of X COS of X or a sin of T COS of T sorry as T approaches Infinity is is ever going to grow okay it reduces faster so this entire thing approaches zero um so we know that I Das of a is equal to Z and then of course we've got to evaluate it at Zer so e to 0 a sin of 0 + cine of 0 okay A 2 + 1 well e to negative uh 0 is just one uh s of 0 is 0 and COS of 0 is uh one of course and so we've get we get I dash of a is = to 0 - 1 / a^ 2 + 1 okay so i- of a then can be written as -1 over a^ 2 + 1 now then um why are we even doing this in the first place um this is because from here we can get back what I of a is the what the integral is uh because that's what we were trying to do originally wasn't it we were we defined I of a to be the integral from 0 to Infinity sin of XX e to netive a x DX um if we can evaluate if we can find out what I of a is equal to we have therefore found out what this is equal to and that is going to massively help us because all we need is this when a is zero to solve our original integral okay so let's go ahead and um find what I of a is well I of a is going to be the integral of both sides because we have the derivative of a i- of a there so to get rid of it uh or to find I of a the original I OFA function we integrate both sides now luckily for us the uh the negative integral of 1 / a^2 + 1 I should say this is with respect to a um is um is is a standard integral okay I of a is therefore equal to negative AR tan of a + C okay now I of a we know to be the integral from 0 to Infinity of sin of x/x e to the ax DX and so we can say this is equal to the negative of R tan of a + C right and so um well we need to we need to find that plus C so we need to um we need to find an a value for which we can sort of know the value of this integral which will allow us to rearrange for C okay and the one that Springs to mind which is why we introduce this negative sign so early on is as a approaches Infinity okay as a approaches Infinity as we discussed e to the ax approaches zero and so when a is infinity we know what this integral is equal to as a approaches infinity or let's say just when a is infinity I know that's not technically correct but when a is is infinity okay well well the integral from 0 to Infinity of sin of x /x * 0 is going to just give us 0 we just got zero and that's equal to negative R tan of infinity plus C well as a approaches infinity negative AR tan of infinity or negative AR tan of a as a approaches Infinity gives us < / 2 okay plus C which implies that C is Pi / 2 when a is um you know approaching Infinity Okay Okay cool so we now know that the integral from 0 to Infinity of sin of x /x uh e to a x DX is equal to the ne AR tan of a plus plus < / 2 and we are almost there our original integral the integral from 0 to Infinity of sin of X /x DX was when I was evaluated or when a was equal to z i evaluated at zero and so we are going to do exactly that okay so a is equal to zero now which means our solution is going to be negative R tan of 0 + piun / 2 negative R tan of 0 is 0+ piun / 2 is just < / 2 which implies then that our integral from 0 to Infinity of sin of X /x DX is equal to pi/ 2 and that is it quite a beautiful solution thank you for watching

Channel: Jago Alexander

Views: 21,825

Keywords: Maths, Maths education, Integration, Calculus, AP Calculus, University maths, Mathematics, Richard Feynman, Feynman technique, A levels, A level maths, Educational content

Id: BqrS75PBCno

Length: 19min 32sec (1172 seconds)

Published: Fri Mar 22 2024