Epic integral solved using Feynman's trick (feat. the digamma function)

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Күн бұрын

Пікірлер: 41

@元兒醬 Ай бұрын

06:25 Gamma(3-t/2) = (1-t)/2*Gamma(1-t/2) and you can use Euler's reflection formula I(t) = (1-t)/4*PI*Csc(t+1/2) it's quicker

@nicolasgobert8324 26 күн бұрын

you can solve it with standart tequnices only😁: First trigsub then split up the log, then use Kingsrule to sum up two of the Integrals, then do power reduction formula with cos²x. After that you will be left with only one Integral, then make u-sub 2x=z. Then you can use ln[cos(x/2)]=-0.5*ln(2)+0.5*ln[cos(x)+1]. And now last steps are easy: do Integration by Parts and solve the last easy Integral and you will get -Pi/4❤‍🔥

@homerthompson416 Жыл бұрын

Always thought it was funny how much Feynman hated contour integration in his book Surely You're Joking. Love his differentiation under the integral sign method but contour integration can be so easy sometimes thanks to the residue theorem.

@renesperb Жыл бұрын

A different approach would be as follows : You first use that 1/(x^2+1)^2 is related to the derivative of a geometric series : -2*x /(x^2+1)^2 = (1/(x^2+1)^2) ' Then use the substitution x= exp t which leads to the integral of t*exp[-3 t]/(1+exp(-2*t))^2 (limits - inf to + inf), which one expands into the derivative of a geometric series.Interchange summation and integration to get a series whose sum you have to identify as -π/4.

@maths_505 Жыл бұрын

Noice

@violintegral Жыл бұрын

When I saw the thumbnail, I first thought of evaluating this integral using integration by parts and symmetry, which I've done before, but when I read the title, I thought of using Feynman's trick in a much different way. Let I(a) be the integral from 0 to infinity of log(x)/(x^2 + a) w.r.t. x, a > 0. With the substitution x = sqrt(a)*u, I(a) is equal to the integral from 0 to infinity of 1/sqrt(a)*log(sqrt(a)*u)/(u^2 + 1) w.r.t. u. Since log(ab) = log(a) + log(b), the integral can be split in two, and the integral from 0 to infinity of log(u)/(u^2 + 1) w.r.t. u is easily shown to be 0 by the substitution u = 1/t. Now all that's left is the integral from 0 to infinity of log(sqrt(a))/sqrt(a)*1/(u^2 + 1) w.r.t. u, which simply evaluates to pi/4*log(a)/sqrt(a) since log(sqrt(a)) = log(a^(1/2)) = 1/2*log(a). Therefore I(a) = pi/4*log(a)/sqrt(a). To find the integral in question, use Feynman's trick and differentiate I(a), giving I'(a) as the integral from 0 to infinity of -log(x)/(x^2 + a)^2 w.r.t. x. From here, it is clear that we want to find -I'(1). Differentiating the other expression for I(a) w.r.t. a gives I'(a) = pi/4*(2 - log(a))/(2a^(3/2)) and -I'(1) = -pi/4. Also, any use of the Digamma function in your method can be avoided since your I(t) = 1/2*Gamma((3 - t)/2)*Gamma((1 + t)/2) can equivalently be expressed as 1/2*Gamma(1 + 1 - (1+t)/2)*Gamma((1 + t)/2) = 1/2*(1 - t)/2*Gamma(1 - (1 + t)/2)*Gamma((1 + t)/2) = 1/2*(1 - t)/2*pi/sin(pi*(1+t)/2) = pi/4*(1-t)/sin(pi/2(1 + t)) since Gamma(z + 1) = z*Gamma(z) and Gamma(z)*Gamma(1 - z) = pi/sin(pi*z). But you got to use that cool Digamma formula by your method so I think your solution is more interesting! Some more integrals featuring the oily macaroni constant and the Digamma function would be great 👌 and just special functions in general, too

@vincentnguyen2558 Жыл бұрын

By Euler's Reflection formula, I believe you can rewrite (1/2) Γ((3-t)/2) Γ((1 + t)/2) as π(1-t)/4 sec(π t/4). This might be an alternate way to differentiate I(t) :)

@homerthompson416 Жыл бұрын

Yeah that's how I did it and just assumed the reflection formula was how he derived I(t) since it was setup so clear by (1/2) Γ((3-t)/2) Γ((1+t)/2) = (1/2) ((1-t)/2) Γ((1-t)/2) Γ((1+t)/2) = ((1-t)/4) Γ(1 - (1+t)/2) Γ((1+t)/2) = (1-t)π/4 csc((1-t)π/2) = (1-t) π/4 sec(πt/2) where we used Γ(x+1) = x Γ(x)

@pluieuwu Жыл бұрын

i was thinking about this as well cos he uses the reflection formula so much xd

@mohan153doshi Жыл бұрын

The Beta function, the gamma function and now the digamma function. This is becoming more and more awesome. Can't wait for what integral is next gonna scream out 'SOS' as 505 sets to annihilate it.

@jieyuenlee1758 9 ай бұрын

Apply fenyman trick on beta function Straight away let u=x² The intergral become Int[0,inf) 1/4 u^(1/2-1)lnu/(1+u)² dx =1/4 dB/du(u=1/2,v=3/2)

@nicogehren6566 Жыл бұрын

very nice approach

@illumexhisoka6181 Жыл бұрын

First time see Feynman trick but not on integral it self (usually we try to find the value of the defined function on certain value not it's derivative)

@Matrix_Architect Жыл бұрын

awesome as always

@manstuckinabox3679 Жыл бұрын

Another great installment in the Feynman technique destroying integrals franchise

@xyzxyzxyzxyz636 Жыл бұрын

Zero is Singularity, not a NUMBER.

@ethannguyen2754 Жыл бұрын

I’m very curious to understand what you mean by this. As I understand it, 0 is an element of the set known as the real numbers, so it is quite strange to me that you say it isn’t a number.

@xyzxyzxyzxyz636 Жыл бұрын

New physics and mathematics. All I can say, is to consider "i" being very REAL and think vectors and geometry. Very important fact is that P=3.22

@ethannguyen2754 Жыл бұрын

@@xyzxyzxyzxyz636 I’m left with more questions and no answers. What is “new physics and mathematics? What does the letter P represent? Why did you change the definition of real numbers to make i a real number? How does any of this make 0 not a number?

@KingGisInDaHouse Жыл бұрын

No zero is a number. It is an integer which makes it a real number. It behaves a lot differently then the numbers around it but it is still a number.

@xyzxyzxyzxyz636 Жыл бұрын

@@ethannguyen2754 "P", is the correct phonetics (pronunciation in Greek language) of "pi", instead of 3.14(), this proportionality constant of nature is 3.22. This affects also the "Fine Structure Constant ", known as 137 or 1/137, e.t.c.