Any calculus book recommendations? Btw are you a mathematician as your profession?

The first substitution is brilliant. I wonder how you came up with it, or if it was the product of trials and errors.

Cool

Do you just make these randomly and solve them, or do you specifically look for nice ones?

Sir , X=1-t/1+t substitution is best for this problem

Very nice evaluation and a very satisfying solution.

i used u=lnx and my last steps are the same as yours

Excellent

Nice.

A very good way to get this sum.

Although one can guess that the original integral must be related to the well known integral of e^(-x^2) , it is quite a long and tricky way to get there.

great work. If you use geometric series right off the bat, you can skip the trig sub.

Nice

Excellent 👌

Don't we have a vertical asymptot at x=pi/2 making the integral diverges ?

what is G constant ??

ridiculously impressive

All you had to do is use the Basel problem result and factor out 1/4 from the even terms and you’d get the answer in 3 minutes lolllll

Nice

Nice

t=1/x dx=-dt/t^2 I=int[1,♾️](t^-2/(1+floor(t)))dt I=int[1,♾️](t^-2/ceil(t))dt I=sum[n=1,♾️](int[n,n+1](t^-2/(n+1))dt) I=sum[n=1,♾️](1/(n+1)•(-1/t)|[n,n+1]) I=-sum[n=1,♾️](1/(n+1)•(1/(n+1)-1/n)) I=1-pi^2/6+sum[n=1,♾️](1/n(n+1)) 1/n(n+1)=1/n-1/(n+1) I=1-pi^2/6+1 I=2-pi^2/6

Damn that's a really nice use of complex numbers and euler's identity

Integration by parts twice then finish with substitutions

They only ask me if it converges or diverges, I hope I will be able to understand how to actually sum it one day!

The given integral is Sum^{\infty}_{0} (-1)^n \int^{1}_{0} dx x^n ln x = Sum^{\infty}_{0} (-1)^n d/dt /t=n \int^{1}_{0} dx x^t = Sum^{\infty}_{0} (-1)^(n+1) 1/(n+1)^2 = Sum^{\infty}_{1} (-1)^n 1/n^2 = - eta(2) = -(\pi )^2/12.

In fact, one could generalize this to I(m,n) = \int^{1}_{0} dx x^m (ln x)^n. Let f(t)= \int^{1}_{0} dx x^t = 1/(t+1). Then, I(m,n) = d^n/dt^n f(t), t=m = (-1)^n n!/(m+1)^(n+1). So, \int^{1}_{0} dx x^m (ln x)^n = (-1)^n n!/(m+1)^(n+1).

Let f(t)= \int^{1}_{0} dx x^t = 1/(t+1). The given integral I = d/dt f(t), t=1 as d/dt x^t = x^t ln x. So, I = -1/(t+1)^2 with t=1 = -1/4.

So, now, how do you prove the Weierstrass product? Start with proving the Eisenstein series for cotangen using the Residue theorem. In fact, that Eisenstein series formula makes even quicker work of this sum.

Awesome technique! I hope you'll make a video discussing on how to evaluate the sum in 15:20

🥰 tysm

Nice and smooth

Cool 😀A DJ taking part in an integral

Int_0^inf u^2/(1+u^4) dx = (pi/5) csc(3pi /5) What did i do wrong?

answer = pi / (2 sqrt(2)) ?

Genial

Very nice! Therefore, the sum from - \infty to +\infty should yield pi coth pi. This is a very useful result for calculating Matsubara sums in finite temperature field theory.

Excellent

Wait dont you need for |x|<1 in order to use the maclaurin series of ln(x+1)

Or just do x=(1-t)/(1+t) just like cipher did and solve it in 3 minutes

Let t=√x > I = 2 \int^1_{0} dt t^3/(t+1) = 2 Sum^{\infty}_{0} (-1)^n \int^1_{0} dt t^(n+3) = 2 Sum^{\infty}_{0} (-1)^n/(n+4) = 2 [ Sum^{\inftyy}_{1} (-1)^n/(n ) + 5/6] = 2[-ln2 + 5/6] = 5/3-2 ln 2.

Integration by parts would be better for this example

i ask the same question (the definite integration ) to the wolfram . wow it give the same answer. actually it also give a indefinite integration form. dosen't know how it can do.

at 9:46 the 2nd term in the summation is not correct. missing (a+1)^2 in denominator.

Marvelous

I let x=y^2 and performed long division.

I let x = cot(y) and integrated by parts twice.

can u do the integral of sqrt(tanx)

never saw such a way of solving a definite integral. pretty cool, thank you

didn't understood 82% of the video, still here to increase the no of comments. Thank Me later