clean integration as always, did not expect the eta function. nice

@maths_505

11 ай бұрын

Wanna know how I came up with it.... I was evaluating a log integral. Made substitution x=e^u and forgot to transform the differential element. Ended up getting a result that didnt agree with wolfram alpha. Retraced my steps, realized the mistake and then completely discarded the original integral in favor of evaluating this one born from a mistake during the original solution development 😂😂😂

@ranjithkumararunachalam3844

10 ай бұрын

@@maths_505 How I learn integration advance, series in calculas......

@ingenuity23-yg4ev

10 ай бұрын

@@maths_505 oh damn thats even cooler lol. sometimes our mistakes lead us to greater wonders as in this case

There's is an easier approach by adding and subtracting x^2*e^x in the numerator followed by integration parts (zero integral). The two integrals you get are textbook results.

Today was a rare day indeed. I looked at the thumbnail of one of your videos and was actually able to do the problem. Cherry on top is that we approached it the same way, cheers for the fun problem!!

You can follow me on Instagram for write ups that come in handy for my videos: instagram.com/maths.505?igshid=MzRlODBiNWFlZA== If you like the videos and would like to support the channel: www.patreon.com/Maths505

This is amazing and smart. Thank you.

Performing the integration of a wild function like this one is pretty much like the saying from Bilbo Baggins: "It's a dangerous business, Frodo, going out of your door. You step into the road, and if you don't keep your feet, there is no knowing where you might be swept off to."

@maths_505

10 ай бұрын

I'm currently rewatching LOTR and I watched fellowship of the ring the day before yesterday 😂

You keep blowing my mind.

No freakin way... I saw the thumbnail and spent 30 (or so) minutes trying to solve this damn thing. I tried a substitution, even a contour integral, all failing miserably. Then I got the idea to find an asymptotic series expansion of (z+1)^(-2) to transform the integral into a series problem. That ended up working out quite well. I basically did exactly what you did in retrospect, and somehow got the answer correct, which blew my mind. Thanks for putting me on a roller coaster.

@maths_505

10 ай бұрын

That's actually the plan with all of my integrals

@SuperSilver316

10 ай бұрын

It might be crazy but I think a contour integral does work. I’m close to developing a solution for this one using a rectangular contour.

Amazing!!!

Love these integrals

Thanks for sharing this beautiful boi. I was playing with other exponents like x^4/(1+exp(x))^3, giving -(7 π^4)/20 + 9 ζ(3) + (45 ζ(5))/2

@maths_505

10 ай бұрын

Beautiful

Nice integral! I wonder...do you have any exotic integrals in the works right now? 🤔

@maths_505

11 ай бұрын

Always bro

The results you used for the series could you provide the link for that.

@maths_505

10 ай бұрын

You mean the eta function? You could just Google it

It would be easier if you multiplied by e^x and e^(-x) and did integration by parts, then you wouldn't need the derivative of the geometric series. It wouldn't be worth it, otherwise I'd do a video on it :) I feel this integral is generated by you in your mind or did you take it from somewhere? ;)

@r2k314

10 ай бұрын

Think about it. This isn't healthy. Please stop.

@erfanmohagheghian707

10 ай бұрын

@@r2k314 what do you mean?!!!! It's actually a skill to be able to develop a tough integral in my mind and then solve it! Who are you to tell me to stop?!

2(1/8-2/27+3/64-4/125+5/216-6/343+7/512...)=2S((-1)^k*(-k/(k+1)^3))=0,15/0,16... Ho completato i calcoli 2(3/4Z(3)-pi^2/12)

seems weird to leave an answer in terms of a function. is there no better way to express the solution here?

Do you have a phd in math?

@maths_505

10 ай бұрын

Nah I just graduated

@johnporter7915

10 ай бұрын

@@maths_505 grats mate

(1+v)⁻² = 1 - 2v + 3v² - 4v³ + ........ for |vl < 1

I think this can also be done by contour integration + "logarithmic feynman technique" (believe it or not) so we dobedo a u-sub, get ln(u)^2/u(1+u)^2, then we define I(s)=1/((u-1)^s*u^2 ) (becareful whilst defining the branch and how it operates, since we'll be using a keyhole contour the positive real line will be avoided regardlessly). now our desired integral is I(1)''. proceeding with the contour integration and differentiating should yeild the result... I guess LOL edit: no it won't work since there are no residues inside our contour... and many other reasons. specifically that the integral """does not converge""" but luckly if we don't use the x-1 sub, we get that our defined I(s) is actually 1/2(sH(s-1/2)- sH(s/2)+1 ) which might lead to something nice. where Hn is the nth harmonic number.

Please diversify a little, because viewers are tired of watching integration, sum, or differentiation

@maths_505

11 ай бұрын

Maybe you're getting tired but the rest of us seem pretty happy. On a side note however.....what kind of "new" content are you looking for?

@ali97x52

11 ай бұрын

@@maths_505 I mean, through this mathematics that you possess, you solve some physics issues in any theory, so that everyone benefits

@maths_505

11 ай бұрын

@@ali97x52 bruh I ain't a physicist....not even a mathematician in the true sense either......but if you hand me a physics problem that results in an integral or a DE (as they so often do), I might be able to solve it. I guess I could try something along the lines of teaching some undergrad physics....but right now I'm working on other things. Thanks for the suggestion though....its actually a nice proposition. But right now the channel is pretty much tough math for fun but I'd love for it to branch out in different ways in the future. But I am never letting go of doing awesome calculus and differential equations just for the thrill of it.

@danielrosado3213

11 ай бұрын

@@maths_505 based

@daddy_myers

11 ай бұрын

​@@ali97x52It IS *Maths* 505 after all...