i solved by doing x = tan(θ) then using identity : 1 - tan²(θ) = 2tan(θ) / tan(2θ) and then split up log then use series expansion for ln(tan(θ)) which can be obtained from from getting the Fourier series for ln(sin(θ)) and ln(cos(θ)) then taking their difference
Ques i found, which seemed impossible Let f(x)=(x⁴)+(x³)+(x²) Let y depict the inverse of the function f(x) Then, evaluate the integral from (lower bound)2A to (upper bound)8A of (1/(y²+y⁴)) with the limit A tends to ♾️infinity The variable of integration is obviously dx, as y will be a function of x only
u=(1-x)/(1+x) substitution twice Yes I would calculate it this way First way is too complicated First step can be ln(1-x^2) = ln(1-x) + ln(1+x) but it will not simplify integral a lot
Well we can't split up the sum using linearity but it should work in the form I've presented where we have a difference of terms. An example of where something like this works is the series representation of the digamma function.
Con semplici calcoli si arriva a I=-(1/2)INT(1+cosθ)dθ..θ=0..π/2...poi non ricordo come si risolve quell integrale..ho provato con β e Γ..ma non porta a niente..ho rifatto i calcoli..I=(-πln2/4)-2((1/2)(-πln2/2+G)=πln2/4-G
yea thats what i did, you will then have to use a series expansion for ln(tan(x)), which can be obtained by getting Fourier series for ln(sin(x)) and ln(cos(x)) then taking their difference
Hi,
"ok, cool" : 1:50 , 5:09 , 6:09 , 7:22 , 10:52 , 11:39 , 10:56 ,
"terribly sorry about that" : 6:37 , 8:58 , 10:46 , 11:56 .
11:39 is terribly sorry about that, 10:56 is nothing
Great job tho
@@biscuit_6081 Ok, thanks
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Great video , thanks Kamaal for your work hard work that you've puten into this and other videos .
Integral bounds jumpscare 1:40
i solved by doing x = tan(θ) then using identity : 1 - tan²(θ) = 2tan(θ) / tan(2θ) and then split up log
then use series expansion for ln(tan(θ)) which can be obtained from from getting the Fourier series for ln(sin(θ)) and ln(cos(θ)) then taking their difference
Thanks for your innovative integrals.
Heyy Brooo fan from Sri Lanka ❤🎉
Greetings from future fields medalist
What are you going to do that will win you the Fields Medal?
@@xinpingdonohoe3978 math
Ques i found, which seemed impossible
Let f(x)=(x⁴)+(x³)+(x²)
Let y depict the inverse of the function f(x)
Then, evaluate the integral from (lower bound)2A to (upper bound)8A of (1/(y²+y⁴)) with the limit A tends to ♾️infinity
The variable of integration is obviously dx, as y will be a function of x only
@davidblauyoutube yeah it must be of that form, cause in the original question it said that the ans is lnα, find α. Do share the solution pls
u=(1-x)/(1+x) substitution twice
Yes I would calculate it this way
First way is too complicated
First step can be ln(1-x^2) = ln(1-x) + ln(1+x)
but it will not simplify integral a lot
Love from india
What about convergence of the series? The sum of 1/k * ln 2 is the harmonic series...
Well we can't split up the sum using linearity but it should work in the form I've presented where we have a difference of terms. An example of where something like this works is the series representation of the digamma function.
Why at 1.40 bounds becomes (0,1) instead remains (0, inf) ?
Because bounds (0, inf) was only by mistake.
Right ! I watched the video again
Con semplici calcoli si arriva a I=-(1/2)INT(1+cosθ)dθ..θ=0..π/2...poi non ricordo come si risolve quell integrale..ho provato con β e Γ..ma non porta a niente..ho rifatto i calcoli..I=(-πln2/4)-2((1/2)(-πln2/2+G)=πln2/4-G
An infinite series of ridiculous digamma terms (and other stuff)? Are you series?! 🤣
isnt it much easier to just split the integral and x=tan(\theta)
yea thats what i did, you will then have to use a series expansion for ln(tan(x)), which can be obtained by getting Fourier series for ln(sin(x)) and ln(cos(x)) then taking their difference
maht
Nice advertising kamaal
First
Please stop OKcooling your videos
@@jpf119 that's the primary reason why almost 60k people have subscribed in the first place.