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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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    $u=\sqrt{n}t$ So $u\in[0,\sqrt{n}]$2012-06-06

1 Answers 1

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