A Monster Integral with Feynman's Integration Trick

hello guys welcome back PK here so in this video we're going to be working on this nice integral using five months technique and I'll be introducing all of the sub techniques related to it so stay tuned okay so our question is evaluate this integral from 0 to 1 then we have 32 * sin to ^ of 5 Ln X over Ln X and DX so I'll be using this U substitution first so let me call this U as now Ln X that means your X is now equal to e to the^ of U so we can say DX is equal to e to the^ of U and du so using this let's rewrite this integral so your I is then going to be equal to integral from now negative Infinity to zero and then we have 32 * s to the^ of 5 U and then we have e to the^ of U over now U U and then du so let's use another substitution let me call this as a v substitution so let me call this v as negative U okay then we can rewrite this integral I as now let me pull this 32 out then we have 32 * integral of positive Infinity to zero and we have S to the^ of five of now V and then times e to the^ of V over now v d of V so that we can rewrite this as now okay so this is the same thing as positive 32 times integral of now then zero to infinity and we have S to the^ five of now we have V time e to the^ of netive V over positive V and then we have DV okay so this has to be the basic setup okay then this is when we can use f men's technique so I'll be calling this capital letter of f now t as inte 32 times integral from 0 to Infinity okay then we have S to the^ five of the V and then times e to the power of TV and over V and DV so what we can know is f Infinity is now equal to zero then we can talk about derivative of this F of t so F Prime T So this frime T has to be equal to now then 32 * integral of 0 to Infinity then we have S now to the^ 5 V * e to the^ of TV and DV okay then we can rewrite this using complex number so this has to be the same thing as 32 times integral from zero to Infinity then we have e to the power of now IV minus E to the^ of i v over now 2 I to the^ of 5 * e to the^ of TV and DV okay so this is what we have so far and then we can rewrite this by pulling first of all one over I out then inside of it I'll be using this binomial right so for this I can rewrite this as let me pull this one over I out okay and we have um integral of from 0 to Infinity then all we makeing this bracket and inside of your bracket first of all we have e to the power of 5 i v minus e to the power of5 i v and then minus using your binomial 54 multiply by now e to the power of so 3 IV minus E to the^ of -3 IV then after this we should have now the plus sign so plus now five now three and then we can multiply this to now e to the power of just IV minus e to the power of netive IV okay and then close your bracket time e to the power of TV and DV okay now if we calculate this then we have 1 over I okay that times now I'll be making a bracket then we should have now 1 over t - 5 i - 1/ t + 5 I okay then we have minus 5 times and inside of it we should have 1 over t - 3 I then minus 1/ t + 3 I close and then plus now we have 10 10 * 1 / t - I and then - 1 / t + I okay then let's calculate this and get your expression so it has to be the same thing as then 1/ I 1 over I times okay the first these two should be combined as now than 10 I over t^ 2 + 25 and then - 5 * 6 I over t² + 9 and then + 10 times now then 2 I over t^ 2 + just one okay so that's why we can cancel those eyes out on your numerator then your F Prime T is just equal to 10/ t^2 + 25 and then now we have -3 over t² + 9 and then we have plus now 20 over t² + 1 and from this what I'm going to use is if you have integral of 1 over t^2 plus say a square DT this has to be the same thing as 1 over a times now arc tangent of t over a I'll be using this okay then using all of this let me fix this capital letter T to be greater than one then we can talk about how your F of T minus F of one that is integral from 1 to T F Prime T DT since we already know F Infinity is equal to zero then we can rewrite everything based on this expression so what we looking for is now Okay negative 10 over 5 now times r tangent of now T over 5 minus AR tangent of 1 over 5 okay then we have plus 30 over 3 times now the AR tangent of t over3 minus arc tangent of 1 over 3 okay then we have minus then 20 * now arc tangent of now T minus arc tangent one okay then we already know when your capital letter T when your capital letter T is going to Infinity then your F of infinity is going to now zero so that's why arc tangent of this T over 5 now this is equal to Pi / 2 okay then using this we can re write so let's rewrite this then we have -2 * R tangent of now then 1 over five and then we have plus now 2 * pi / 2 okay then after we have 10 times so plus 10 times R tangent of now then 1 over 3 and then we have -10 * pi over 2 lastly we have -20 * AR tangent of 1 and then plus 20 * pi / 2 okay then making sure this arc tangent of the one arc tangent of one this is equal to pi over 4 so that's why we can rewrite this again then if you rewrite this we should have now -2 time AR tangent of Now 1 over 5 okay then we have plus pi and we have + 10 * AR tangent of 1 / 3 and - 5 pi and again we have -20 R tangent of 1 but -2 times now pi over 4 and plus now we have 10 pi since -20 * Pi 4 this is equal to5 Pi so that's why we can also rewrite this okay so we have Pius 5 Pi that is4 pi4 Pi minus 5 Pi is9 Pi + 10 pi is just equal to Pi so we have pi and then -2 * R tangent of 1 over 5 and lastly we have plus 10 * R tangent of 1 over three okay this is the exact value of the I the integral that we are looking for so final answer for this question is this expression okay so pretty interesting integral using five men's technique so I'll be back in more videos for more questions like this sometime soon

Channel: PK Math

Views: 731

Keywords: feynman's integration trick, PkM&#%, feynman, feynman's trick, feynman integration technique, feynman integration trick, challenging integral, integral, calculus, evaluating challenging integral, complex analysis, pk math, yt:cc=on, feynman's intergration trick, feynmans intergration trick, feynmans integration trick

Id: 1ekw6dg1HWQ

Length: 10min 21sec (621 seconds)

Published: Mon Feb 12 2024