Integration using the gamma function

Рет қаралды 10,386

Prime Newtons

Күн бұрын

In this video I used the gamma function to evaluate a definite integral that would otherwise, be very hard to evaluate.

Пікірлер: 35

@bridgeon7502 4 ай бұрын

Interesting how pi always shows up in integrals 🤔

@user-ky9kv5je9s 4 ай бұрын

I think you make a mistake when you put Gamma(-1/2+1) at the end of video. You might say Gamma(1/2+1)

@PrimeNewtons 4 ай бұрын

Oops 🙊 you're right.

@Heemashti 4 ай бұрын

Please include your logarithms tutorial, for example log(log(logx))

@hosseinmortazavi7903 15 күн бұрын

Nice teacher

@marcolima89 4 ай бұрын

Your handwriting is seriously one of the prettiest I have ever seen. just one minor detail, in the end result, the 27 in the denominator should be +/- 27. Thanks a lot for these videos, amazing quality. As a mechanical engineer, I miss sometimes these math lessons.

@datboy038 Ай бұрын

Should it? He wasnt solving for x and the square root always gives out the absolute value

@curtpiazza1688 4 ай бұрын

WOW! Great STUFF! 😊

@williammartin4416 4 ай бұрын

Thanks!

@PrimeNewtons 4 ай бұрын

Thanks a lot

@emmanuelonah4596 4 ай бұрын

Very interesting

@jennymissen3523 4 ай бұрын

How groovy is that?!

@xinpingdonohoe3978 4 ай бұрын

In my opinion, I prefer to just use the actual factorial notation. Some people say "that's only defined for nonnegative integers" but I disagree. The Riemann zeta function, through analytic continuation, can be defined almost everywhere. Not all of it will satisfy the initial definition of ζ(x)=sum n≥1 of 1/n^x, but we still call it ζ. ζ(-1) is well defined and equal to -1/12, even if it's not the actual sum n≥1 of n. It's the same here. The factorial function has been analytically continued to all but countably many numbers. And whilst most don't satisfy the original definition of k!, the product n=1 to k of n, it's still the factorial function. (1/2)! is well defined and equal to √π/2, even if it's not the product from n=1 to 1/2 of n.

@adw1z 4 ай бұрын

The factorial function n! is only defined on the non-negative integers n (0,1,2,...), and technically speaking the gamma/pi function is not an analytic continuation of the factorial, but rather an interpolation of the factorial function, as such an extension to the complex plane C is not unique (analytic continuation (AC) applies only to connected domains D, which is not the case here with the non-negative integers - as such, any AC must be unique). The gamma function is what we use by convention in place of this non-unique extension via the integral definition (satisfies gamma(z+1) = zgamma(z), gamma(1) = 1) which indeed only converges for Re(z) > 0. But it is not unique: for example, the so-called pseudogamma function also successfully interpolates the factorial. Hence, denoting something such as the equivalent of gamma(3/2) as (1/2)! can be ambiguous. The AC only applies to the gamma function itself, taking the domain to Re(z) < 0 (on which it is meromorphic, poles at non-positive integers); this is unique by the identity theorem, and this time our domains Di (from Re(z)>0, continued to slices -1 < Re(z) < 0, -2 < Re(z) < 0 etc... until ultimately onto Re(z)

@parthasarathy4990 2 ай бұрын

Errata - it is gamma(3/2) = 1/2 * gamma(1/2)= root pi/2 .... Not gamma (-half+1)

@sovietwizard1620 Ай бұрын

Even from the answer we can tell that the error function is involved in the indefinite integral answer to this.

@Mutlauch 4 ай бұрын

Hi, would't the Pi-function have a z-1 in the exponent of t, because you substituted z -> z-1 with respect of the Gamma-funciotn? Greetings from Germany :)

@lawrencejelsma8118 4 ай бұрын

He did something strange like evaluating π(Z) = √π/2 results instead. I was just as confused but just accepting that the "Indefinite integral" from 0 to infinity of √te^(-t)dt evaluates to √π/2 after substitution of those limits. Hopefully the ending part is correct! 😂