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

Integration by parts with u = ln(x^3+1) , dv = 1/x^3 dx \int_{0}^{\infty}\frac{1}{x^3+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{1}{x^3+1}dx + \frac{1}{2}\int_{0}^{\infty}\frac{1}{x^3+1}dx \int_{0}^{\infty}\frac{1}{x^3+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{1}{x^3+1}dx + \frac{1}{2}\int_{\infty}^{0}\frac{1}{\frac{1}{u^3}+1}(-\frac{1}{u^2})du \int_{0}^{\infty}\frac{1}{x^3+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{1}{x^3+1}dx + \frac{1}{2}\int_{0}^{\infty}\frac{u^3}{1+u^3}\frac{1}{u^2}du \int_{0}^{\infty}\frac{1}{x^3+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{1}{x^3+1}dx + \frac{1}{2}\int_{0}^{\infty}\frac{u}{1+u^3}du \int_{0}^{\infty}\frac{1}{x^3+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{1+u}{1+u^3}du \int_{0}^{\infty}\frac{1}{x^3+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2-u+1}du \int_{0}^{\infty}\frac{1}{x^3+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{1}{\left(u-\frac{1}{2} ight)^2+\frac{3}{4}}du u-\frac{1}{2} = \frac{\sqrt{3}}{2}w

high class. It's very difficult for me to solve this.

6:05 Isn't arctan(cot(@)) just pi/2-@?

I think you can make these videos shorter and better if you combine some of the obvious steps. But good video otherwise.

Apologies for the disruptive sound towards the end, had to cut the video short.

Very cool solution. You can't pull the y terms all the way out of a double integral, can you?

Very cool indeed

1+cos(x) = 2cos^2(x/2) then integration by parts with u = x , dv = 1/(2cos^2(x/2))dx Finally substitution u = cos(x/2)

Can you make a video explaining the Last function you mentioned.

From where you have integrals to calculate ? I have one \int_{\theta}^{\pi}{\frac{\sin{\left(\left(n+\frac{1}{2} ight)t ight)}}{\sqrt{2\left(\cos{\left(\theta ight)} - \cos{\left(\t ight)} ight)}}dt} Hint using trigonometric identities and integration by parts derive recurrence relation and dont forget base cases for recurrence

Integration by parts with u = arctan(x) , dv = 1/(x+1)dx Substitution u = (1-x)/(1+x)