The Proof is Trivial!

Pretty much. A more direct way would be to differentiate the answer to problem 22, which of course is implicitly what you did.

Problem 294**/***

Prove
$\displaystyle \frac{\zeta^2(2)}{2}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^4+m^2n^2}$

Problem 295***

Suppose we have a closed contour $\gamma$ in the complex plane. Then it defines an enclosed area. We shall define

$S = \dfrac{1}{2i}\displaystyle\int_{\gamma}\bar z\ dz$

To be the signed area enclosed by $\gamma$

Show that S is real, and that it recovers $\pi r^2$ for a circle, show both signs can occur, and explain the significance of the sign.

Solution 295

Consider the parametrization $\gamma(t) = r e^{it}$ for $t \in [0, 2 \pi]$ which is a circle of radius r about the original. Using Cauchy's integral theorem, we know that this contour will give a general result for $\gamma$.

Integrating in the anticlockwise direction:

Hence $\displaystyle S = \frac{1}{2i} \int_{\gamma}\bar z\ dz = \frac{1}{2i} \int_0^{2 \pi} (r e^{-it})(ri e^{it}) \ dt = \frac{r^2}{2} \left[ t \right]_0^{2 \pi} = \pi r^2 \ \square$

Integration clockwise along this contour would return the value of $- \pi r^2$.

You didn't show S was real..? (I did mean for any smooth gamma!)

CIT implies the contour we choose does not affect our result so long as the turning number is the same, is that right?

And an 'enclosed area' sort of implies either a simple smooth loop either clockwise or anticlockwise (real in both cases) .

anyone can join this coven, but we insist that it remain finite…

Consider a smooth manifold $X$ with an open coven…

Solution 271

Now, using the fact that $\displaystyle \frac{1}{\Gamma(z)}= \exp(\gamma z)z \prod \left(1+\frac{z}{n} \right)e^{-\frac{z}{n}}$

Solution 272

From the above result, we obtain $\displaystyle \gamma = -4\int_{0}^{\infty} e^{-x^{2}}x \ln x dx$.

We also have $\displaystyle \digamma(z) = \int_{0}^{\infty} \left(\frac{1}{e^{x}x}- \frac{1}{e^{zx}(1-e^{-x})} \right) dx$ (follows directly from the definition)

which gives $\displaystyle \gamma = \int_{0}^{\infty} \left( \frac{1}{xe^{x}}-\frac{1}{e^{x}-1} \right) dx$.

Considering $\displaystyle \int_{0}^{\infty} \left( \frac{1}{e^{x}-1}-\frac{1}{x(1+x^{k})} \right) dx=0$, we get the fifth representation.

Solution 273

$\begin{aligned} \displaystyle \int_{0}^{1}\int_{0}^{1} \frac{1-x}{(1-xy)(-\ln xy)^{-z}}dx dy & \overset{y \mapsto \frac{y}{x}}= \int_{0}^{1}\int_{0}^{x} \frac{1-x}{x(1-y)(-\ln y)^{-z}}dydx \\&= \int_{0}^{1} \frac{y- \ln y -1}{(1-y)(-\ln y)^{-z}}dy \\&= \Gamma(z+2) \zeta(z+2) -\Gamma(z+1) \end{aligned}$.
Now, letting $z \to -1$, we get $\displaystyle \int_{0}^{1}\int_{0}^{1} \frac{x-1}{(1-xy)\ln xy}dxdy= -\gamma$.

Solution 275

Some complex analysis shows that $\displaystyle \int_{0}^{1} \frac{1}{(1-xy)x(\ln^{2}(\frac{1-x}{x})+\pi^{2})}dx = \frac{1}{\ln(1-y)}+\frac{1}{y}$

Oh how I love waking up to random problems like this! I'm wondering if you found the same nice 'trick' that I did (the method is reminiscent of Fubini's Theorem for integrals).

Solution 294

Changing the order of the series, changing that bases WLOG, adding and then separating:

$\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{m^2(m^2+n^2)}= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{n^2(m^2+n^2)} \Rightarrow \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{m^4+m^2n^2}$
$\displaystyle =\frac{1}{2} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{m^2+n^2} \left( \frac{1}{m^2}+\frac{1}{n^2} \right) = \frac{1}{2} \left( \sum_{m=1}^{\infty}\frac{1}{m^2} \right) \left( \sum_{n=1}^{\infty}\frac{1}{n^2} \right) = \frac{\zeta^2 (2)}{2} \ \square$

This is incomplete. The point of this question is to prove (or not assume) any non-*/** theorems. 'Felix Felicis' produced a similar proof but deleted it when he realized that he had to prove the theorems.

None of these verbal statements are sufficient for mathematical proof.

Looks right, but you're going to need to include some more details.. I mean, who in their right mind could follow that without pen and paper?

The integral actually began as a conjecture in 2004 (though it was immediately solved since it's not that hard) but the mathematician who came up with it was unable to prove it!

Also, it is not true for all complex z so either you have made a mistake or that's another crucial point you have left out.

Some Galois Representations shows that $\displaystyle x^n+y^n \not= z^n \forall n >2 \land x,y,z,n \in \mathbb{N} \land x, y, z \not=0$... you're going to either need to show your working or give the name of the theorem you are using. All you are essentially doing is telling us you can do it?

Sorry to sound like such a dick man, but if anyone besides you posted those solutions, people would probably think they didn't actually do it given the lack of detail.. also the satisfaction is reading someone's solution is in understanding something, or seeing how someone else did something.

Sorry to sound like such a dick man, but if anyone besides you posted those solutions, people would probably think they didn't actually do it given the lack of detail.. also the satisfaction is reading someone's solution is in understanding something, or seeing how someone else did something.

And, just by the way, if you think that you will always be able to read mathematical text without a pen and paper, then, I dare say, you are completely wrong. Moreover, when the author omits the details, that one who benefits is the reader.

Solution 294-5

Don't want to clog up the thread:

Hmm, a page with comments is actually a really bad design for a forum. It should be a directed graph - usually it would be a tree, although you could draw branches together in a single reply. You'd then be able to reply to a comment by creating a child node to it, and quote people by making the original post a parent. Then a single conversation could be tracked really easily - just follow the arrows, the first-created arrow first (that is, weight the arcs by when they were formed).

It sounds a lot like you want reddit