I want to solve the following integral exactly
\begin{align*}
\int_{0}^{\pi}\frac{\cos(t^2)}{1-2a\cos(t)+a^2}dt,\quad -1
Exact value of Integral
1
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integration
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0$a$ lies between $-1$ and $ 1$. i.e., $-1
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1Is the exponent in the numerator on the cosine or on the argument? – 2017-02-07
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0Due to the [Poisson kernel][1] we have: $$ \int_{0}^{\pi}\frac{\cos(t^2)}{1-2a\cos t+a^2}\,dt = \frac{1}{1-a^2}\sum_{n\in\mathbb{Z}}a^{|n|}\int_{0}^{\pi}\cos(\theta^2)e^{in\theta}\,d\theta \tag{1}$$ and if we expand $\cos(\theta^2)$ as a Fourier series we certainly end with a series whose terms depend on Fresnel integrals. [1]: https://en.wikipedia.org/wiki/Poisson_kernel – 2017-02-07
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0Thanks....I read this link.......:) – 2017-02-07