But What's Feynman's Trick All About?

hello guys welcome back PK here so in this video we're going to be working on this interesting integral using Fan's technique so the question is evaluate this integral from 0 to one then we have four times s square of Ln X over Ln of 1 /x then DX first of all let's rewrite this so I'll be rewriting this integral I as now okay equal to um integral from 0 to 1 we have 4 * s now square of Ln X still and then let me rewrite this denominator as then negative Ln of just DX and we have DX first of all we'll be working on the U substitution right so I'll be calling this U as this Ln X that means your X is equal to e to the^ of U that also means your DX is equal to e to the^ of uu and du okay so using this let's rewrite this right so integral is then integral is then going to be equal to okay so we have the -4 right so4 integral from negative Infinity to zero then we have S square of the U times e to the^ of U over now U and then we have du and from this we will be working on another substitution let me call this as V substitution so your V is now let me call your v as uh netive U okay so using this we can rewrite this integral s then now ne4 now times integral of infinity to zero and we have now then sign sare of V * e to the^ of V over V of D of V so that's why we can rewrite this integral again as now I is is equal to positive4 of integral from 0 to Infinity okay then we have S square of V * e the power of V over V and DV okay and this is where we can use Fan's technique right so using Fan's Technique we can call your say capital letter of the T right is just going to be now then four times integral of 0 to Infinity then we have sin sare of the V * e the^ of TV over V and DV this is going to be defined and all positive value of the key okay so that's why we already know that your F of one capital letter of f of1 is this integral I and then F capital letter of infinity is equal to zero okay then from this we can talk about your derivative of capital letter of f of T right so we can talk about F Prime of T this is going to be just equal to the4 integral of 0 to Infinity then we have S square of V time just the E to the^ of TV and DV then we can rewrite this using complex number right so this is the same thing as then4 * integral of 0 to Infinity then we can rewrite the sin square of the V part right as now e to the power of i v minus E to the^ of i v over 2 I okay then we should Square this and then we can multiply e to the power of TV and we have DV evv okay then based on this expression we can just work this out right if you work this out then it has to be um integral from zero to Infinity okay then evaluating this then it has to be then your e to the power of okay say it has to be 2 i v and then okay minus 2 and then plus e to the^ of the -2 i v time e to the^ of TV okay and we have DV and if you work this integral out then we should have e to the power of okay 2 i parenthesis 2 i - t * V over this 2 I minus t okay then the second term is now equal to -2 * e the^ of - TB over - t and then the third term is now plus e to the^ of 2 I minus t * V over that's -2 I minus t okay then it should be from zero to infinity and if you work this out then the first term is equal to 1 over T minus 2 I and the second term is now the -2 over T and then the last term is plus one over t + 2 I so that's why if you combine these two fractions right then it has to be 2 T over T square + 4 and then of course we have -2 over T okay this is what we have okay based on this works say capital letter T that is greater than one so for say capital letter T that is greater than one we already know your capital letter F of T minus capital letter F of 1 this is your integral from one to this capital letter T of derivative of your capital letter F of T and DT okay so that is why now we can work on your integral this has to be the same thing as integral from one to T and then you should have this 2 T over t + 4 minus 2 over T and we have DT this is what we need to evaluate right and if you work this out then it has to be just equal to Ln of now t^ 2 + 4 minus 2 * Ln of t Okay from one to this capital letter of the T so that that's why if you evaluate this then we should have now then Ln this is equal to Ln of capital letter t² + 4 um over t² minus now we have Ln of five and based on this if you let your T the capital letter T goes to Infinity so let your capital letter T goes to Infinity then your capital letter F of infinity is is now equal to zero so that's what we can say limit capital letter T is going to Infinity then for this term right so Ln of T ^2 + 4 over t^ 2 this is the same thing as just Ln of one which is now equal to zero okay so based on this we can say your capital letter F of one is then going to be this Ln of five so that's why what we need to evaluate for this integral is just equal to Ln 5 so the answer for this question is L and five it's a pretty interesting integral using five months technique so I'll be back in more more questions like this sometime soon

Channel: PK Math

Views: 512

Keywords: PkM&#%, feynman's integration technique, feynman trick, feynman's integration trick, integral, integration, challenging integral, how to use feynman's integration technique, calculus, feynman, pk math, yt:cc=on

Id: c4uqiFNXsbg

Length: 8min 19sec (499 seconds)

Published: Sat Feb 03 2024