Basel Problem Integral: Solving with Calculus

[kairama15]

Summary:: Using an integral and taylor series to prove the Basel Problem

The Basel problem is a famous math problem. It asked, 'What is the sum of 1/n^2 from n=1 to infinity?'. The solution is pi^2/6. Most proofs are somewhat convoluted. I'm attempting to solve it using calculus.

I notice on wolframalpha.com that the integral of -ln(1-x)/x dx from 0 to 1 is pi^2/6. I also noticed that its taylor series evaluated at 1 is the sum of x^n/n^2 from n=1 to infinity. This is a link between pi^2/6 and the infinite sum of inverse squares.

If I can evaluate this integral, it can prove the Basel problem using just undergraduate calculus. I tried substitutions of u=ln(1-x), u=1-x, 1-x=e^u and even trig substitutions like x=sin(x). I'm having trouble evaluating it. Any ideas?

 

[fresh_42]

One proof uses $\int_0^1\int_0^1\frac{1}{1-xy}dxdy$ and evaluates it by two different methods: as geometric series $\sum (xy)^n$ and by substitution $u=\frac{y+x}{2}\, , \,v=\frac{y-x}{2}\,.$

 

[kairama15]

Yes, that double integral is related to my integral. If you evaluate the inner dx from 0 to 1 you get my integral.

I'm having trouble with the final logarithmic integral I mentioned.

 

[fresh_42]

Yes, that double integral is related to my integral. If you evaluate the inner dx from 0 to 1 you get my integral.

I'm having trouble with the final logarithmic integral I mentioned.

Use the substitution instead. That gives you an arcus tangent, rather than a logarithm.

And maybe you could write your formulas in LaTeX. I have no nerves to decipher it hidden in a wall of text.
See https://www.physicsforums.com/help/latexhelp/

 

[fresh_42]

We already know that the result is $\frac{\pi^2}{6}$. This means that we need some trigonometry, either a trigonometric function or polar coordinates. The real logarithm doesn't provide that.

 

[suremarc]

The fact that $\int_0^1 -\frac{\ln(1-x)}{x}\\ dx=\frac{\pi^2}{6}$ is essentially a result of formally manipulating the Taylor series of the integrand and integrating termwise. The integral is equal to $\mathrm{Li}_2(1)$, the dilogarithm at 1, which is a special case of the polylogarithm identity $\mathrm{Li}_s(1)=\zeta(s)$.

Anyway, I’m skeptical that this would lead to anything without great effort, since this is basically just a series of formal manipulations. You still need something more.

 

[kairama15]

Use the substitution instead. That gives you an arcus tangent, rather than a logarithm.

I can't use Latex because I'm on mobile and there's no 'backslash' key.

Thanks for the advice!
I was able to prove that the

double integral of dx*dy/(1-(xy)^2)

written as the infinite series you mentioned does indeed become the

sum of 1/n^2 from n=1 to infinity

However I am having trouble solving the double integral to prove it is equal to pi^2/6.

If I make the substitutions you recommended, I get:

double integral of 4*du*dv/(1-u^2+v^2) from u=0 to 1 and v=0 to 1.

Integrating with respect to du and evaluating from 0 to 1, I get the remaining single variable integral:

integral of 4*atan(1/sqrt(-v^2-1))/sqrt(-v^2-1) from v=0 to 1.

Neither me nor wolframalpha can solve this integral using basic analytical functions.

I keep getting stuck evaluating this double integral. Any ideas?