@amWhy I did what I could do , marking the relevant comments , close-voting the question and downvoting question and answers. Because of the unjustified many upvotes, we won't be ablr to remove this post without moderator help. The undeletion is sad and shows why this site simply does not work.

@user21820 only D8 needs one more delete-vote.

CD

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Is (x^2-1)/(x-1) continuous at x=1? has been asked and answered many times. My first suggested duplicate target is not the best choice however, if you decide to vote as a duplicate, better choose one from my other comment.

@BillDubuque Hi. I recently read a comment from your side saying that "Integration problems usually succumb to a standard bag of tricks so require little ingenuity." Well, I have to say I'd love to be your apprentice. Every time I looked at the problem below I have no idea on how to go $$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt=\frac{\pi(3-\gamma)}{32}$$

It's from here mathoverflow.net/q/57944.

Have a good day!

@amWhy @JoséCarlosSantos or any golden badge holder of calculus: If you have the time could you have a look at math.stackexchange.com/q/4319134/42969? It was closed as a duplicate, but not with the best dup target (which was my fault). Better targets are pointed out in the comments.

@user97357329 In case that was not meant as a joke, do note that I wrote "usually...". And also be aware that one can go even further and show that integration is undecidable in general. But this has little to do with the matter at hand - just like your example.

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@MartinR Done.

@JoséCarlosSantos Great, thank you.

Who on earth upvoted this ? One more delete-vote needed.

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Hi, any easy proof for my root-problem ? Also, I found an interesting new question from another user.

@Peter No, not really unfortunately, but that's because I couldn't spend enough time on it. Let me know the other question.

this

I checked the powers of $5$ upto $5^{30}$ and I always found a number with the desired property.

The polynomial with the coefficients $1,2,3,\cdots ,n$ is irreducible upto $n=2\ 000$. This special case should be possible to be solved completely.

@Peter I think this along with what is written there, answers the question, right?

Not quite , although this is an interesting partial result ! But the author wants it for every odd number.

I think the rest of the attempt covers every other number, right?

@TeresaLisbon To be honest, I do not see it. How can we use the solution for , say , $5^{100}$ to get it for , say $5^{100}\cdot 2773$ ?

@Peter much appreciated.

@Peter I'll comment on the answer post now, we have gone off-topic.

@TeresaLisbon I agree.

@TeresaLisbon Is this proof correct ? I can only approve that the result is correct.

Too little, too late, @TheSimpliFire. Nothing needing you here now.

Darn, I missed the chance to say hello to @Pedro. I'll just sent a hello to him now. \o

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CD. The answer is no better there.

I do still consider, out of consideration for other users, so as not to post a comment to anyone planning on being gone for a week, or weeks at a time, that both users and mods leave rooms when leaving rooms. It is one source of confusion that mods (and users, cuz we are different, apparently) can ameliorate to leave chats, when one leaves a chatroom. Note, to leave a chat, without logging off the site, below the tags for this room, there are three links available. The first (left) is...

..."leave". Just click it! (This is not necessary for anyone planning to be back, say, the following day.) Cheers to all! Just trying to reduce confusion and wasted time spent by users trying to ping another user/mod. ;D