The result should be π/4, you dropped a factor of ½ in the IBP section. Also, how about just performing a magical transformation straight from x to -x, and seeing where that takes you? You'll be pleasantly surprised.
@@GreenMeansGOF After you simplify your transformed integral, you'll be left with log(1+e^-x), which can be re-written as log((1+e^x)/(e^x)). Just apply properties of logarithms, and you'll notice that you're left with your original integral and some other basic Calc 2 integral.
I was an integration by parts hater, until I met this integral... bro this was so awesome! although I dought that I would have been able to guess a factorization of e^x/2... thanks for teaching me this trick!
Your solution is more elegant than mine. I IBP like you did but then I brute forced the resultant integral by splitting the integral at 0 and using geometric series for 1/(1+exp(-x)). Had to derive some other general integrals so it took me almost 2 whiteboards to get to the answer XD.
I think a way to conceptually understand this is by a "limiting integral". We have I = S_-oo^oo x*ln(1+e^x)/(1+x^2)^2 dx Now what is interesting about this integral is that if you take the limit x --> +oo, we can define this to be, since e^x >> 1, IL ~= S_-oo^oo x*ln(e^x)/(1+x^2)^2 dx = S_-oo^oo x*x/(1+x^2)^2 dx = S_-oo^oo [x/(1+x^2)^2]*x dx which we then integrate by parts Differentiate x Integrate [x/(1+x^2)^2]dx --> -1/2(1+x^2)^2 and in the first limit we get IL ~= 0 - (-1/2)*S_-oo^oo dx/(1+x^2)^2 = 1/2 * (pi/2) = pi/4 which is exactly our result. So using the integral version of the squeeze theorem, if IL = lim(I as x-> oo) = pi/4 then so does I at the bounds x --> +-oo since the integral is symmetric.
Integration by parts + odd even decomposition of function odd even decomposition of function Function can be decomposed into f(x) = (f(x)-f(-x))/2 + (f(x)+f(-x))/2 and (f(x)-f(-x))/2 is always odd and (f(x)+f(-x))/2 is always even Finally we get integral 1/2Int(1/(1+x^2),x=0..infinity)
The answer is pi/4
I forgot a factor of 1/2 during the integration by parts
i thought i was wrong 😂, just let t=-x, simplify it. Then there’s going to be some easy integrations. Don’t even need a pen to write.
There is a much simpler method. Making substitution x -> -x reduces the integration to elementary
Yes
The result should be π/4, you dropped a factor of ½ in the IBP section.
Also, how about just performing a magical transformation straight from x to -x, and seeing where that takes you?
You'll be pleasantly surprised.
Damnnnnn that is awesome!
Sorry, I don’t see it. The log terms combine but I do not know what to do from there.
@@GreenMeansGOF After you simplify your transformed integral, you'll be left with log(1+e^-x), which can be re-written as log((1+e^x)/(e^x)). Just apply properties of logarithms, and you'll notice that you're left with your original integral and some other basic Calc 2 integral.
@@daddy_myers oh! I see.
This is a real ingenious solution. I enjoyed the video👍
I was an integration by parts hater, until I met this integral...
bro this was so awesome! although I dought that I would have been able to guess a factorization of e^x/2... thanks for teaching me this trick!
SUIIIIIIIIIIIIIIII
Your solution is more elegant than mine. I IBP like you did but then I brute forced the resultant integral by splitting the integral at 0 and using geometric series for 1/(1+exp(-x)). Had to derive some other general integrals so it took me almost 2 whiteboards to get to the answer XD.
Points for the effort
I think a way to conceptually understand this is by a "limiting integral".
We have
I = S_-oo^oo x*ln(1+e^x)/(1+x^2)^2 dx
Now what is interesting about this integral is that if you take the limit x --> +oo, we can define this to be, since e^x >> 1,
IL ~= S_-oo^oo x*ln(e^x)/(1+x^2)^2 dx = S_-oo^oo x*x/(1+x^2)^2 dx = S_-oo^oo [x/(1+x^2)^2]*x dx
which we then integrate by parts
Differentiate x
Integrate [x/(1+x^2)^2]dx --> -1/2(1+x^2)^2
and in the first limit we get
IL ~= 0 - (-1/2)*S_-oo^oo dx/(1+x^2)^2 = 1/2 * (pi/2) = pi/4 which is exactly our result.
So using the integral version of the squeeze theorem, if IL = lim(I as x-> oo) = pi/4 then so does I at the bounds x --> +-oo since the integral is symmetric.
That was amazing, thank you very much.
awsome !
This is not annihilation but nuking. Good one
DESTROYING INTEGRALS WOOOO!
SUIIIIIIIIIIIIIIII
Holy. Shit! That was awesome!
What program do you use for your videos? I love the way you can scroll right down.
It's just the default software on my note phone
@@maths_505 cool, thanks!
Amazing
Integration by parts + odd even decomposition of function
odd even decomposition of function
Function can be decomposed into
f(x) = (f(x)-f(-x))/2 + (f(x)+f(-x))/2
and (f(x)-f(-x))/2 is always odd and (f(x)+f(-x))/2 is always even
Finally we get integral 1/2Int(1/(1+x^2),x=0..infinity)
Can anyone suggest me a book to start with feynmanns integrals???
There's no such book....however whenever you see an integral, try going feynman on it....that's how you practice
I = int (-inf, +inf) xln(1+e^x) /(1+x²)²dx
Let, x=tant=>dx=sec²tdt
I= int[-π/2, π/2] sintcost ln(1+e^tant) dt
=int[-π/2, π/2]-sintcost ln(1+e^-tant) dt
=int[-π/2, π/2] [-sintcost ln(1+e^tant) +sin²t ]dt
=>2I =int[-π/2, π/2]sin²tdt
2I =2[0, π/2]sin²tdt
=2[0, π/2]Cos²tdt
=>2I = [0, π/2](sin²t+cos²t)dt
=π/2
I=π/4
Nice one
I think there should be -1/2(1+x^2) at 0:41 , If I am wrong rectify me....
You're right...I forgot about that factor 🤣
Misalkan y=1+x^2 maka x dx=(1/2)d(1+x^2)=(dy/2).
|=|ln(1+e^x).y^-2(dy/2)
Lol that's beautiful
Bro! been a while
How's it going???
@@maths_505 oh damn, you remembered me! That's crazy haha
Thanks. I'm doing fine :D just haven't been doing math lately xD
@@maths_505 what about you, how are you ?
@@yudoball doing good bro
Just hit 11k subs alhamdullilah so looks like I'm doing something right 😂
@@maths_505 that's nice. And you had a couple videos that blew up with +100k views.
Stay healthy and take care of yourself bro :D.
The result should be pi/4
Monsters don't exist.
i think answer should me pi/4 can u please check once
Yup I forgot the 1/2 on the way
1/2ln2+1/4pi....no,correggo ln2 non c'è
Maths 420
Wtf
your pens are distracting and presentation & writing very cluttered
Why isn’t this integral Odd?