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This is from an old qualifying examination question.

Let $a>1$ be fixed. Show that

$$ \displaystyle A_N=\pi i a \int_1^N t^{a-\frac{3}{2}}e^{\pi i t^a} dt $$

converges to some complex number as $N \rightarrow \infty$.

I tried using integration by parts, but no luck. Is there something that I'm missing here?. Any hints and comments are greatly appreciated.

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I can barely see it but what is $s$? –  1015 Jan 9 '13 at 1:45
    
@julien: the term inside the integral is exp ^ {pi i t^a}. There is no s anywhere. –  Jack Dawkins Jan 9 '13 at 1:48
    
On my screen, there is also $t^{\alpha-s/2}$. But I can not either write @user54755 at the beginning of my comment. I might need a new computer. –  1015 Jan 9 '13 at 1:53
    
@julien: Sorry about that inside the integral there is t^{a-3/2} times what I wrte before. I wrote that in \LaTeX code I see no reason why it does not show properly. Thanks for looking though. –  Jack Dawkins Jan 9 '13 at 1:56
    
Ah! Ok. I swear this $3$ really looks like an $s$ on y screen. Sorry. –  1015 Jan 9 '13 at 1:58

1 Answer 1

up vote 3 down vote accepted

It looks like an integration by parts should do the job, though.

Setting $u=t^{-1/2}$ and $dv=t^{a-1}e^{i\pi t^a}dt$, you get: $$ A_N= N^{-1/2}e^{i\pi N^a} +1 + \int_1^N \frac{1}{2} t^{-3/2}e^{i\pi t^a}dt. $$ Now it is clear that this converges to $$ 1+\int_1^{+\infty} \frac{1}{2} t^{-3/2}e^{i\pi t^a}dt. $$ In particular, note that the last integral converges absolutely since $3/2>1$.

Edit: I have just seen your last comment. For the first term, we have $|N^{-1/2}e^{i\pi N^a}|=1/\sqrt{N}\rightarrow 0$ as $N\rightarrow+\infty$.

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Thanks so much for your answer. I see what went wrong now. I totally forgot that $e^{i\pi}$ terms are unimodular and therefore the integral converges. –  Jack Dawkins Jan 9 '13 at 2:51
    
@user54755 You're most welcome. Plus you had already got the good idea. –  1015 Jan 9 '13 at 3:25

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