This integral will improve your advanced math skills

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Күн бұрын

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@Tosi31415 9 ай бұрын

amazing solution using many branches of maths at the same time, it's crazy how you can come up with these at this rate (please don't stop)

@_DD_15 9 ай бұрын

Love your content. Really good stuff!! No time wasted, straight to the point! Great channel!

@maths_505 9 ай бұрын

Thank you my friend

@Noam_.Menashe 9 ай бұрын

I once solved this integral but with the (sin/x*lnx) squared. Ever since then I can't look at derivatives of gamma functions the same.

@Okay_adityaa 9 ай бұрын

I am learning so many new concepts just from your Integration videos.

@maths_505 9 ай бұрын

Integrals are fun this way. Solving them often forces you to learn more tools and concepts from other branches of math.

@Mathematician6124 9 ай бұрын

Hey bro, is it similar to your solution?? I did it before on my own this way. Let I(a) = intgrl e^-ax sinx lnx /x dx from 0 to infinity I(a-> infinity) =0 I'(a) = - intgrl e^-ax sinx lnx dx from 0 to infinity Now let's work on, S= intgrl e^-ax sinx lnx dx from 0 to infinity Now prepare for integration by parts, take sinxlnx to be differentiated and take e^-ax to be integrated. So, it becomes, 1/a intgrl e^-ax cosx lnx dx from 0 to infinity + 1/a intgrl e^-ax sinx dx from 0 to infinity 1/(2a) intgrl e^-ax (e^ix + e^-ix) lnx dx from 0 to infinity + 1/a arctan(1/a) 1/2a intgrl (e^-(a-i)x + e^-(a+i)x) lnx dx from 0 to infinity + 1/a arctan(1/a) ..........( 1 ) Now think of intgrl (e^-(a-i)x + e^-(a+i)x) lnx dx from 0 to infinity Now let's take J(b) = intgrl x^b ((e^-(a-i)x + e^-(a+i)x) dx from 0 to infinity J'(0) = intgrl (e^-(a-i)x + e^-(a+i)x) lnx dx from 0 to infinity J(b) = T(b+1)/(a-i)^(b+1) + T(b+1)/(a+i)^(b+1) We use T'(1)= - gamma J'(0)= - ln(a-i) /(a-i) - ln(a+i)/(a+i) - gamma/(a-i) - gamma/(a+i) Use a-i= sqrt(a^2 +1) e^(-i arctan(1/a)) and a+i=sqrt(a^2 +1) e^(i arctan(1/a)), to evaluate those ln() things and simplyfy Thus Evaluate J'(0), we will get J'(0)= {- aln(a^2 +1) - 2arctan(1/a) - 2a(gamma)} /(a^2 +1) Put it in.... (1) and then we finally get, S= - {ln(a^2 +1) - 2a arctan(1/a) + 2gamma} /(2(a^2 +1)) So, I'(a) ={ln(a^2 +1) - 2a arctan(1/a) + 2gamma} /(2(a^2 +1)) Integrate both sides from infinity to 1, here integration by parts of a arctan(1/a) /(a^2 +1) da will be enough, and thus we get I(1)= - pi/8 * ln2 - pi/4 *gamma

@maths_505 9 ай бұрын

Similar but yours is better mashallah

@Mathematician6124 8 ай бұрын

@@maths_505Hey Friend, would you please solve this integral for me?? I'm stuck. intgrl ln(1+x) ln(1+x^2) /(1+x) dx from 0 to 1 Edited :I solved it sorry bro to disturb you. But you can make a vedio on this 😊

@edmundwoolliams1240 9 ай бұрын

Yeah! Beautiful result involving E-M, Pi, and ln2 😊

@MikeB-q8v 8 ай бұрын

Hi, you might consider that sin(x)/x = integral(0,1)cos(a*x)da this removes the issue with the constant of integration and does not require taking the imaginary part of a logarithmically divergent expression. Of course you got the right answer!

@taterpun6211 9 ай бұрын

I love clever tricks like ignoring the c because it's real; mathematicians should be as lazy as possible. Here's a solution using the laplace integral trick ∫(0,∞)fgdx=∫(0,∞)L(f)L^-1(g)dx you showed in a previous video: ∫(0,∞)sinxlnxe^-x/xdx =-γ∫(0,∞)sinxe^-x/xdx-∫(0,∞)((-γ-lnx)/x)sinxe^-xdx (apply trick to both integrals, L(sinxe^-x)=1/((x+1)^2+1)) = -γ∫(0,∞)1/((x+1)^2+1)dx-∫(0,∞)lnx/((x+1)^2+1)dx = -γ∫(1,∞)1/(x^2+1)dx-∫(1,∞)ln(x-1)/(x^2+1)dx =-γπ/4+∫(0,1)lnx/(x^2+1)dx-∫(0,1)ln(1-x)/(x^2+1)dx =-γπ/4-G-∫(0,1)ln(1-x)/(x^2+1)dx For the last integral, let I = ∫(0,1)ln(1-x)/(x^2+1)dx and J = ∫(0,1)ln(x+1)/(x^2+1)dx. using substitution x = (1-u)/(u+1) I = ∫(0,1)ln(2x/(x+1))/(x^2+1)dx = ln2∫(0,1)1/(x^2+1)dx+∫(0,1)lnx/(x^2+1)dx-J I+J = πln2/4-G I-J = ∫(0,1)ln((1-x)/(x+1))/(x^2+1)dx = ∫(0,1)lnx/(x^2+1)dx = -G I = πln2/8-G Hence ∫(0,∞)sinxlnxe^-x/xdx = -γπ/4-πln2/8 You could also just use the same laplace parameterization and property L(f/t)=∫(s,∞)L(f)(s')ds', but that's essentially your solution.

@anasharere 9 ай бұрын

❤

@CM63_France 9 ай бұрын

Hi, "ok, cool" : 2:03 , 2:32 , 4:25 , 5:58 , 8:31 , 8:47 , "terribly sorry about that" : 2:42 , 8:45 .

@wohargRadu 8 ай бұрын

This would be nice if you can switch derivation and integration but for that the integral must converge. As Exp(-x).sin(x) . ln(x) /x ~1.1. ln(x) = ln(x) when x->0 it would have been correct to say that the integral of ln (x) converges in the neighbourhood of 0 what justifies the switch because the integral converges trivially in + infinity.

@Calcprof 9 ай бұрын

The part about ignoring C is fun.

@maths_505 9 ай бұрын

Indeed

@MrWael1970 9 ай бұрын

Thank you for your stunning solution. It is very interesting.

@satyam-isical 9 ай бұрын

1:38 If it looks that things are getting worser,then we are on the right track ~Boi Kamaal😂😂

@pcrlillie 9 ай бұрын

What a great journey of integral…

@giuseppemalaguti435 9 ай бұрын

A 4:30 si può usare u=ax,senza usare la trasformata di Laplace..

@trelosyiaellinika 9 ай бұрын

Beautiful!

@maths_505 9 ай бұрын

Indeed

@박선호-t3f 9 ай бұрын

how can i say Im(ln(1-i))=-pi/4 rather than 2n*pi-pi/4, and use its actual value to the integral? what happens when feynman's trick is extended to complex number?

@maths_505 9 ай бұрын

We only need the principle value.

@onusiddartha1641 9 ай бұрын

Good as expected of u

@bnice24 9 ай бұрын

Calculus destroyer😮😊

@threstytorres4306 9 ай бұрын

Why does W.A says that the integral has the answer of (-π/8)(2¥ + ln2)?; ¥ = Euler-mascheroni constant

@Mathematician6124 9 ай бұрын

ln(sqrt2) =1/2 ln(2) So

@threstytorres4306 9 ай бұрын

@@Mathematician6124 Oh I have just realized that he writes Ln(√2) instead of Ln(2) my bad 😅

@Haxislive766 9 ай бұрын

Can you explain euler mascharoni constant gamma please?

@carultch 9 ай бұрын

Consider the harmonic series, the sum of 1/k, from k=1 to n. This is a series that diverges, but that diverges very slowly. Try plotting the terms of this series in Excel. The shape of the curve should look familiar, as it is very similar to ln(x). Even applying a logarithmic trendline, you'll see that it does a damn good job at matching this series. In the long run, at extremely large values of this series, there is a fixed difference between ln(x) and the harmonic series. That difference is the Euler-Mascheroni constant.

@anasharere 9 ай бұрын

i wanna ask yoy something if you dont mind i tried to solve this using laplac but i faced with laplac (lnx/x) when i put it in wolframe I suprised because there was 2 answer laplac(lnx/x) excite but integral 0 to inf of (e^-sx lnx/x)= convegre !! also the constant C the you ignored in the video = γ²/2 +π²/12 How is that ?? how lim x→∞ of 1/2lnx²+γlnx is equal to γ²/2 +π²/12 sorry for the long question

@y4ssin345 9 ай бұрын

Great explanation! but why is it enough to just check for the case alpha > 0 to determine that Im(c) = 0?

@maths_505 9 ай бұрын

Well it's the constant of integration so it's independent of alpha. Positive alpha just made it stick out like a sore thumb.

@y4ssin345 9 ай бұрын

Ohh I see. I completely forgot about it being a constant my bad. Thanks alot

@maths_505 9 ай бұрын

@@y4ssin345 all good mate. Happens to all of us. Often we'll forget the entire thing completely 😂

@Haxislive766 9 ай бұрын

😂😂

@energeticgorilla 9 ай бұрын

what software/app do you use for these videos? its very pretty and i would love to use it myself 😊

@maths_505 9 ай бұрын

Samsung notes

@karelvanderwalt3625 9 ай бұрын

@@maths_505 Came with the Tablet ?

@maths_505 9 ай бұрын

@@karelvanderwalt3625 yes it's the default software on all Samsung tablets.

@mathalysisworld 9 ай бұрын

Nostalgia❤❤

@GudduBind-v2y 2 ай бұрын

Infinity ♾️=2

@A.Hisham86 9 ай бұрын

I didn't understand the imaginary part trick at the beginning.

@bandishrupnath3721 9 ай бұрын

The poor C will be remembered.

@maths_505 9 ай бұрын

Or forgotten.....

@cameronspalding9792 8 ай бұрын

What you have defined as I(alpha), I would have called J(alpha)

@seesky112 9 ай бұрын

sinx=?*e^ix , how does it become like this?

@SuperSilver316 9 ай бұрын

I’ve been encountering this problem. One where Feynman’s trick is resulting in crazy divergence. One thing I think that remedies this problem is integrating the expression with respect to your integration parameter while you are in the middle of the setup of your anti derivative. In this case here’s what I mean (I’m gonna use v for the E-M constant) I’(a) = (v+ln(a))/(a) 1/a = int(exp(-ax)) from 0 to inf Let’s use that here I’(a) = int((v+ln(a))*exp(-a*x)) from 0 to inf Now integrate with respect to a, and interchange the order of integration I(a) = int(v*exp(-ax)/x+Ei(-a*x)/x-exp(-a*x)*ln(x)/x) from 0 to inf + C Here I have performed the integration over a, and invoked a special function. I think we like to avoid these, but the nature of the problem maybe dictates its purpose. Now just take the limit as a goes inf, C = 0 pops out rather trivially, as I(inf) is also equivalently zero. Then we can go back to our derivative and proceed as usual with the Laplace transform definition we had. This is at least a framework for how I would consider the problem of solving for the constant. I didn’t really prove you could interchange the order of the integration like that and then take the limits under the integral sign, but it does work so maybe that’s something.

@BladimirRemon 9 ай бұрын

Chevere 😊

@Mathematician6124 9 ай бұрын

Oh Remon how are you?? You must be doing good. Do you remember me??You praised me here for that solution kzbin.info/www/bejne/n52wonabeLR4jqMsi=TMpRAQSEqq-ypZeM