Complex Analysis: A Cute Integral

[@user-wu8yq1rb9t]

Finally ... The King of Complex Analysis on KZbin Is Here . I was watting for long time (I missssssssseeeed you) !

[@Decrupt]

Complex analysis ftw!

[@sergpodolnii3962]

This episode is a banana)

[@user-injective]

Welcome back

[@AliAkl-un2ys]

Hi can you solve zeta(3)

[@Anonymous-Indian..2003]

Life is very complex......... We have to study mathematics for understanding it

[@frankargenti]

around min 20 .... you cannot say from -pi/4 to pi/4 --- that implies that you have already solved the limit over that section of integral -You can ofc but you need to prove the continuity of I[f(theta)]

[@gileadlevy9055]

King!

[@rayrash1]

(-i \pi/4)^2 = - \pi^2/16 and not \pi^2/16. This changes your final result to - \pi^3/48.

[@Anonymous-Indian..2003]

Mah bro came back 🎉

[@SussySusan-lf6fk]

Hey😊 I can solve this problem without complex analysis....... Though it's big

[@Anonymous-Indian..2003]

@@SussySusan-lf6fk Are we stalking each other ? 😂 We catch eat other on every math platform 😅

[@Anonymous-Indian..2003]

@@SussySusan-lf6fk Can you send me the solution how you solved without complex analysis ?

[@SussySusan-lf6fk]

@@Anonymous-Indian..2003 I'm commenting in the main comments section.... I may comment tomorrow too... It's big... I've to write and edit first

[@SussySusan-lf6fk]

@@Anonymous-Indian..2003 wait my friend, I'll share it tomorrow for sure... It's 70% done....

[@froggieperi5534]

sorry but you are the real cutie here :-)

[@SussySusan-lf6fk]

Int - 1 to 1, arctanx ln((1+x^2)/2) /(1+x) dx int 0 to 1, arctanx ln((1+x^2)/2) /(1+x) dx - int 0 to 1, arctanx ln((1+x^2)/2) /(1-x) dx I - J I= int 0 to 1, arctanx ln((1+x^2) /2) /(1+x) dx let x=(1-t) /(1+t) I= int 0 to 1, pi/4 ln(1+t^2) /(1+t) dt - int 0 to 1, pi/2 ln(1+t) /(1+t) dt - I + int 0 to 1, 2arctant ln((1+t)/2) /(1+t) dt 2I = 3pi (ln2)^2 /16 - pi^3 /192 - pi (ln2)^2 /4 + int 0 to 1, 2arctant ln(1+t) /(1+t) dt For the integral, integrate by parts, we'll get 2I = pi (ln2)^2 /16 - pi^3 /192 - int 0 to 1, ln^2 (1+t) /(1+t) dt Now do the J J = int 0 to 1, arctanx ln((1+x^2)/2) /(1-x) dx Use x=(1-t) /(1+t) J= int 0 to 1, (pi/4 - arctant) ln{(1+t^2)/(1+t)^2} /{t(1+t)} dt J= int 0 to 1, (pi/4 - arctant) ln{(1+t^2)/(1+t)^2} /t dt - I I'm stepping up some simple steps ahead J = pi^3 /96 - pi^3 /24 - int 0 to 1, arctanx ln{(1+t^2)/(1+t)^2} /t dt [let this be S] - I Now let's solve the integral S S =int 0 to 1, arctant /t ln{(1+t^2) /(1-t)^2} dt + int 0 to 1, 2 arctant /t ln{(1-t)/(1+t)} dt = pi^3 /16 - int 0 to 1, pi lnv /(1-v^2) dv + int 0 to 1, 4 lnv arctanv /(1-v^2) dv [after taking (1-t)/(1+t) =v] Now For the main integral we had I - J = 2I - pi^3 /96 + pi^2 /24 + pi^3 /16 - pi^3 /8 + int 0 to 1, 4lnx arctanx /(1-x^2) dx = pi/16 (ln2)^2 - pi^3 /192 - int 0 to 1, ln^2 (1+x) /(1+x^2) dx - pi^3 /32 - int 0 to 1, 4lnx arctanx /(1-x^2) dx {we transformed them into one variable x} Now we'll use these results from a source .... I'll give the link int 0 to 1, ln^2 (1+x) /(1+x^2) dx = 7 pi^3 /64 - 2Gln2 + 3 pi (ln2)^2 /16 - 4Im(Li3{(1+i)/2}) int 0 to 1, lnx arctanx /(1-x^2) dx = - 5 pi^3 /128 + G/2 ln2 - pi (ln2)^2 /32 + Im(Li3{(1+i)/2}) So we have at last, pi^2 (ln2)^2 /16 - pi^3 /192 - pi^3 /32 - pi^2 (ln2)^2 /16 + 3 pi^3 /64 = pi^3 /64 - pi^3 /192 = pi^3 /96 drive.google.com/file/d/1jFyFMH4VPBcxCRkvvyq1dOiFO_0T-cJG/view?usp=sharing See page 3

[@SussySusan-lf6fk]

Now I used some results, int 0 to 1, arctanx ln{(1+x^2)/(1-x)^2} /x dx = pi^3 /16 We'll use, 2 int 0 to inf , lnt /((t+1)^2 + x^2) dt = ln(1+x^2) arctanx /x int 0 to 1, arctanx ln(1+x^2) /x dx - int 0 to 1, 2arctanx ln(1-x) /x dx =int 0 to 1, 2 int 0 to inf lnt /{ (t+1)^2 + x^2} dt dx - 2 int 0 to 1, ln(1-x) arctanx /x dx = 2 int 0 to inf, lnt arctan(1/(1+t) dt - 2 int 0 to 1, arctanx ln(1-x) /x dx Now take 1/(1+t) = z int 0 to 1, 2 arctanz ln{(1-z)/z} /z dz - int 0 to 1, 2 ln(1-x) arctanx / x dx = - int 0 to 1, 2 lnx arctanx / x Using tanx series, = pi^3 /16 And int 0 to 1, ln(1+x^2) /(1+x) = 3/4 (ln2) ^2 - pi^2 /48 It's easy if you use direct Feynman technique

[@Anonymous-Indian..2003]

@@SussySusan-lf6fk Hey, the drive link which you've published has the solutions from the youtube channel "AKSS". Right ?

[@SussySusan-lf6fk]

@@Anonymous-Indian..2003 yes.... I am a fan of his problems Isn't my solution good?

[@SussySusan-lf6fk]

I can solve this problem without complex analysis.... Is anyone willing to see?