How to prove $$\int_{0}^{\infty }\cos \left (x \right )\sin \left (x^{2} \right )dx=\frac{1}{2}\sqrt{\frac{\pi }{2}}\left ( \cos\frac{1}{4}-\sin\frac{1}{4} \right )$$ any hint ?Thanks!
Find $\int_{0}^{\infty }\cos \left (x \right )\sin \left (x^{2} \right )\mathrm{d}x$
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calculus
integration
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2why a question that shows zero effort gets this attention? Close vote! – 2017-01-14
2 Answers
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\begin{align*} \int_{0}^{\infty }\cos x\sin x^2\, \mathrm{d}x&=\frac{1}{4}\int_{-\infty}^{\infty }\left [ \sin\left ( x^{2}+x \right )+\sin\left ( x^{2}-x \right ) \right ]\mathrm{d}x\tag1\\ &=\frac{1}{4}\int_{-\infty}^{\infty }\left [ \sin\left ( \left ( x+\frac{1}{2} \right )^{2}-\frac{1}{4} \right )+\sin\left ( \left ( x-\frac{1}{2} \right )^{2}-\frac{1}{4} \right ) \right ]\mathrm{d}x\\ &=\frac{1}{2}\int_{-\infty}^{\infty }\sin\left ( x^{2}-\frac{1}{4} \right )\mathrm{d}x\\ &=\frac{1}{2}\cos\frac{1}{4}\int_{-\infty}^{\infty }\sin x^{2}\, \mathrm{d}x-\frac{1}{2}\sin\frac{1}{4}\int_{-\infty}^{\infty }\cos x^2\, \mathrm{d}x\tag2 \end{align*} The use the Fresnel integral , we will get the answer as wanted.
$(1)$ : $\sin\alpha \cos\beta =\dfrac{1}{2}\left [ \sin\left ( \alpha +\beta \right )+\sin\left ( \alpha -\beta \right ) \right ]$
$(2)$ : $\sin\left ( \alpha -\beta \right )=\sin\alpha \cos\beta -\cos\alpha \sin\beta $
Use the same way we can get $$\int_{0}^{\infty }\cos x\cos x^2\, \mathrm{d}x=\frac{1}{2}\sqrt{\frac{\pi }{2}}\left ( \cos\frac{1}{4}+\sin\frac{1}{4} \right )$$
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0Nice solution! Thanks. – 2017-01-14
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0Between the second and third lines, I would have liked to see $\int_0^{\infty} = \int_{-1/2}^{\infty} + \int_{1/2}^{\infty}$ and then showing that the parts from $1/2$ and $-1/2$ cancel out. – 2017-01-14
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I thought it might be instructive to present a way forward that exploits Euler's formula. To the end, we proceed.
We begin by exploiting the even symmetry of the integrand and write
$$\begin{align} \int_0^\infty \cos(x)\sin(x^2)\,dx&=\frac12\int_{-\infty}^\infty \cos(x)\sin(x^2)\,dx\\\\ &=\frac12\text{Im}\left(\int_{-\infty}^\infty \cos(x)e^{ix^2}\,dx\right)\\\\ &=\frac14\text{Im}\left(\int_{-\infty}^\infty (e^{ix}+e^{-ix})\,e^{ix^2}\,dx\right)\\\\ &=\frac14\text{Im}\left(\int_{-\infty}^\infty (e^{i(x^2+x)}+e^{i(x^2-x)})\,dx\right)\\\\ &=\frac14\text{Im}\left(2e^{-i/4}\int_{-\infty}^\infty e^{ix^2}\,dx\right)\\\\ &=\frac12\text{Im}\left(2e^{-i/4}\sqrt{\frac{\pi}{2}}(1+i)\right)\\\\ &=\sqrt{\frac{\pi}{2}}\left(\cos(1/4)-\sin(1/4)\right) \end{align}$$