@Arkamis "integral solution" means finding an antiderivative - your answer. I think it is just fine. I would suggest an edit to "How do we know a function doesn't have an antiderivative in closed form?" or something similar.
I mean one answer is to say that smart mathematicians have defined a function to be $$ \operatorname{Si}(z) = \int_0^z \frac{\sin x}{x} \,\mathrm{d}x $$
And when smart mathematicians does that, it usualy means the integral can not be written in an elementary form.
I can never get something in such concrete form. Its swimming around changing all the time, and everytime I write something, a month later, I am not happy with what I have written. :-/
But I actually found a solution that works for me.
I only work when I feel a gist of motivation, if not then I do not work. And I only take small chips at a time. Solving a few problems one day, then might come back a motnth later and do some more.
@N3buchadnezzar and a suggestion of mine is changing $l$ to $\ell,$ particularly in pages 94-95. I think it is nicer, and there is a thread here about this, but I couldn't find.
@N3buchadnezzar didn't you think to publish your book? I'd like to publish a book of limits and integrals. Every day tens of nice things come to my mind. I'd like to have a book if possible ... (maybe one day)
@N3buchadnezzar in page 17, you wrote $\displaystyle\int\frac{1+x^4}{1+x^4}\,dx$, based on what comes afterwards, I guess you meant $\displaystyle\int\frac{1+x^2}{1+x^4}\,dx.$
@N3buchadnezzar if I had a book I'd call it "A collection of 20,000 amazing integrals and limits". That "20,000 " is also related to the nice novel "20,000 Leagues Under the Sea" by Jules Verne.
@N3buchadnezzar ok, it is because I got confused. There are $\mathrm{ln}$'s floating around, but also $\log$ when it seems clear you meant $\log_e$. Example: $\displaystyle \int_0^{e^\pi}\sin\left(\log(x)\right)\,dx$ is quite harder if $\log=\log_{10}$.
@N3buchadnezzar here is a new born (maybe you like it) and it ask for an elementary solution $$\lim_{n\to\infty} n^{1/120}\left(\sqrt{1+\sqrt[3]{1+\sqrt[4]{1+\sqrt[5]{n+1}}}}-\sqrt[120]{n+1}\right)$$
I just created it, and now I'm working on the generalization case.
@robjohn I did like $$\sqrt[4]{1+\sqrt[5]{1+n}}=\sqrt[20]{1+n}\sqrt[4]{1+\frac{1}{\sqrt[5]{1+n}}}= \sqrt[20]{1+n} \left(1+\frac{1}{4\sqrt[5]{1+n}}\right)$$ which gets messier and messier.
though I never got around it to I thought it would be a pain to differentiate between the cases when the starting index of the power series was at 1 or 0
@Charlie Hi...And hi Peter. Sorry if I sounded snippy. I just checked the OP's profile after suspecting that the comments below that particular question were aiming above the level of the OP.
$$\text{ But since,}$$ $$\frac{1}{m^4+m^2n^2}=\frac{1}{n^2m^2}-\frac{1}{n^4+n^2m^2}$$ $$\text{ We get,}$$ $$\frac{\zeta(2)^2}{2}=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{m^4+m^2n^2}$$
@Charlie Good, good. I learned today I only need radiation treatment (learned yesterday about a tumor, but today, assured that the tumor is benign, given the results of a biopsy, but in some cases (10%), leaving intact can lead to becoming cancerous. So very relieved, and am fine with radiation for the next few weeks.
The technique for solving the double sum by noting that $$\frac{1}{m^4+m^2n^2}=\frac{1}{n^2m^2}-\frac{1}{n^4+n^2m^2}$$, was so nice it gave off some other identities like
@amWhy [Referring to your answer to the question about (ab)c=(ac)b] I am not sure that is an explanation of why you need both. Rather, it says that both are sufficient.
@Charlie Absolutely no need for alarm. The tumor is of the sort that usually goes unnoticed, most often never causes trouble, and mine was found earlier than is usual. So I've really no greater risk of cancer, after radiation.
If $f$ is an arithmetic function with $f(d)\ne 0$ for all divisors $d$ of $n$, then $$\sum_{d\mid n} \frac{1}{f(d)^2+f(d)f(\frac{n}{d})}=\frac{1}{2}\sum_{d\mid n}\frac{1}{f(d)f(\frac{n}{d})}$$
@PeterTamaroff I'm really not sure. The phrase seems to come up a lot in the literature for asymptotics for orthogonal polynomials, but I'm having a hard time seeing what the difference between "strong" and normal asymptotics is.
I will take Linear Algebra next "quad"mester. I can also take Analysis II; but I have chosen to sign up for a optional course on sequences and series (I think I mentioned this before). The syllabus of LI is very extensive, and since I am not really well versed in Linear Algebra, I think it might be good to take dedicate a good time of the next four months to this and that little course, and then continue with Analysis, which I am OK with, really.
