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I would like to find the sum of the series

$$ \sum u_{n}$$

where $$ u_{n}=(-1)^n\int_0^1 \cos(nt^2)\mathrm dt$$

Using the change of variable $t\rightarrow \sqrt{n}t$:

$$ u_{n}=\frac{(-1)^n}{\sqrt{n}} \int_0^{\sqrt{n}} \cos(t^2)\mathrm dt\sim_{n\rightarrow \infty} \frac{(-1)^n}{2}\sqrt{\frac{\pi}{2n}}$$

So $\sum u_{n}$ is convergent.

What about

$$ \sum_{n=0}^{\infty} (-1)^n\int_0^1 \cos(nt^2)\mathrm dt$$


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Careful with equivalents when $u_n$ hasn't a constant sign: for example, when $u_n=\log\left(1+\frac{(-1)^n}{\sqrt n}\right)$. However, you can use the fact that $\{\int_0^{\sqrt n}\cos(t^2)dt\}$ is convergent. –  Davide Giraudo Jun 6 '12 at 12:04
I don't get your last question: what with what? You did it above, didn't you? –  DonAntonio Jun 6 '12 at 12:05
$u=\sqrt{n}t$ So $u\in[0,\sqrt{n}]$ –  Chon Jun 6 '12 at 12:35
@Chon Ah, vision problem due to bleary eyed morning. I saw $\sqrt{n}$ as $\pi$. –  Thomas Andrews Jun 6 '12 at 13:05
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1 Answer

up vote 6 down vote accepted

The m-th partial sum of your sum is the real part of $$ \sum_{n=0}^{m} (-1)^n \int^1_0 \exp(int^2) dt = \int^1_0 \sum_{n=0}^{m} \left(-\exp(it^2)\right)^n dt= \int^1_0 \frac{1-(-\exp(it^2))^{m+1}}{1+ \exp(it^2) } dt.$$

The integral $\displaystyle I_m = \int^1_0 \frac{\exp(imt^2)}{1+\exp(it^2)} dt $ tends to $0$ by the Riemann Lebesgue Lemma. Thus $$\sum_{n=0}^{\infty} (-1)^n \int^1_0 \exp(int^2) dt = \int^1_0 \frac{1}{1+\exp(it^2)} dt.$$

We can compute $$ \Re\left(\frac{1}{1+\exp(it^2)}\right) = \frac{1+\cos(t^2)}{\sin^2(t^2)+(\cos(t^2)+1)^2} =1/2$$

so $$\sum_{n=0}^{\infty} (-1)^n \int^1_0 \cos(nt^2) dt = 1/2.$$

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Nice proof, thanks! –  Chon Jun 6 '12 at 12:45
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