DESTROYING ONE SAVAGE INTEGRAL

Рет қаралды 7,991

11 ай бұрын

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Пікірлер: 28

@maths_505 11 ай бұрын

My new channel for formal math courses: youtube.com/@TheHouseOfMathness505?si=ID1g_2aTc2JNAIZD If you like the videos and would like to support the channel: www.patreon.com/Maths505 You can follow me on Instagram for write ups that come in handy for my videos: instagram.com/maths.505?igshid=MzRlODBiNWFlZA== My LinkedIn: www.linkedin.com/in/kamaal-mirza-86b380252

@joniiithan 10 ай бұрын

Bro I said it and I’ll say it again: don’t stop producing your videos. Insane content

@maths_505 10 ай бұрын

#hustle #neverstop😎😂 I'm gonna be dropping videos on both channels alternatively. Today's the next lecture on complex analysis.

@aryaghahremani9304 10 ай бұрын

the 1-t in the denominator had me waiting for the geomteric series the whole video lol

@MrWael1970 10 ай бұрын

Smart Substitutions to get the result. Thank you for your stunning effort.

@xl000 10 ай бұрын

Feynman would destroy all of you in hand to hand combat. He actually served in the special ops during WW II, under a pseudonym

@manstuckinabox3679 10 ай бұрын

bro I should really learn how to invoke these special functions, I did everything the way you did except I completed the square instead of multiplying up and down by (1-u)... amazing conent as always friend.

@insouciantFox 10 ай бұрын

You can actually use some dilogarithm identities to crunch this integral down quickly, but the answer you get isn't as elegant as yours. It also has a pet peeve of mine which is involving complex numbers in a real valued integral.

@holyshit922 10 ай бұрын

I would start in the same way , problem is that interval of integration is [0,1] instead of [0,infinity] I did not get that i can multiply numerator and denominator by 1- u After this I would probably use geometric seres expansion

@lmaorofl3229 11 ай бұрын

yess

@jasonlin5884 10 ай бұрын

Wow ! Too hard to understand. here is an integral question, maybe easier than this. But I can not work out , is there somebody help me ? ( it is derived from gravity of a hollow sphere) integrated function f(x) = ( 1+ux)/ ( (1+u^2+2ux)^(3/2) ) u is a constant 0

@zunaidparker 9 ай бұрын

At 1:34 I was seriously expecting a u-->1/u substitution. I wonder if that would have worked.

@bjorncarlsson787 10 ай бұрын

One could also expand with (1-x²) directly and then substitute x^6 = u ? I didn't even know what the trigamma function was before.

@A2431A 10 ай бұрын

Bro forgot by-parts exist 💀

@vascomanteigas9433 11 ай бұрын

Maxima gives the result with dilogarithms, but using complex arguments: (%i*li[2]((sqrt(3)*%i+3)/2))/(4*sqrt(3))-(%i*li[2](-(sqrt(3)*%i-3)/2))/(4*sqrt(3))+(%pi*log(12))/(2*3^(3/2))-(%pi*log(4))/(2*3^(3/2))

@borhen-di6ik 10 ай бұрын

Hello Sir, what about Matrice Lesson

@daddy_myers 11 ай бұрын

Nice one; however, is there not a closed-form for the result?

@maths_505 11 ай бұрын

The book of Saint Gigachad states in chapter "never skip leg day bro" that all closed forms are beautiful and need to be respected. That includes closed forms in the shape of polygamma functions. It's kinda like those (Γ(1/4))² results......

@daddy_myers 11 ай бұрын

@@maths_505 Coulda said it's tedious, but you coped instead.

@maths_505 11 ай бұрын

@@daddy_myers 😂😂😂 nah buddy I'm serious. If there was a plus sign instead of minus we could've invoked a property of the trigamma function.

@Ramsey_erdos 10 ай бұрын

i know clearly my answer is wrong but have no idea how let I(a)=int xln(ax)/(x^4+x^2+1) I'(a)=1/a int x/(x^4+x^2+1) u+x^2+1 i'(a)=1/2a int from 2 to 1 of 1/u^2-u+1 therefore I'(a)=pi/3root(3)a therefore I(a)=pi/3root3 *lna I(1)=0

@oom_boudewijns6920 10 ай бұрын

If u go from I' to I u forget + c. To calfulate the c u need a value for I(a) where a is something. This is impossible since 0 gives no solution and 1 is the integral itself. U would need a complex 'a' which would, if computable, be something with the trigamma funxtion

@Ramsey_erdos 10 ай бұрын

thank you so much most times ive been doing this i guess c was trivially 0 but this makes complete sense. idk what trigamma function really is so clearly above my level :( what wasted time lololol @@oom_boudewijns6920