CAUCHY SCHLÖMILCH COSINE INTEGRAL

Lets proof the following result:


\int_{-\infty}^{\infty} \cos \left(x^{2}+\frac{a^{2}}{x^{2}}\right) d x=\sqrt{\frac{\pi}{2}}(\cos 2 a-\sin 2 a)



\begin{aligned} \int_{-\infty}^{\infty} \cos \left(x^{2}+\frac{a^{2}}{x^{2}}\right) d x &=\int_{-\infty}^{\infty} \cos \left[\left(x-\frac{a}{x}\right)^{2}+2 a\right] d x \\ &=\cos (2 a) \int_{-\infty}^{\infty} \cos \left[\left(x-\frac{a}{x}\right)^{2}\right] d x-\sin (2 a) \int_{-\infty}^{\infty} \sin \left[\left(x-\frac{a}{x}\right)^{2}\right] d x \\ &=2 \cos (2 a) \int_{0}^{\infty} \cos \left[\left(x-\frac{a}{x}\right)^{2}\right] d x-2 \sin (2 a) \int_{0}^{\infty} \sin \left[\left(x-\frac{a}{x}\right)^{2}\right] dx\\ &=2 \cos (2 a) \int_{0}^{\infty} \cos \left(x^{2}\right) d x-2 \sin (2 a) \int_{0}^{\infty} \sin \left(x^{2}\right) d x \\ &=2 \cos (2 a) \sqrt{\frac{\pi}{8}}-2 \sin (2 a) \sqrt{\frac{\pi}{8}} \\ &=\sqrt{\frac{\pi}{2}}(\cos 2 a-\sin 2 a) \qquad \blacksquare \end{aligned}


The Cauchy Schlömilch transformation (proof can be found here)


\int_{0}^{\infty} f\left(\left(a x-\frac{b}{x}\right)^{2}\right) d x=\frac{1}{a} \int_{0}^{\infty} f\left(x^{2}\right) d x


The following results hold:


\begin{aligned} &\int_{0}^{\infty} \sin x^{a} d x=\sin\left( \frac{\pi}{2 a} \right) \Gamma\left(1+\frac{1}{a}\right) \\ &\int_{0}^{\infty} \cos x^{a} d x=\cos \left( \frac{\pi}{2 a} \right) \Gamma\left(1+\frac{1}{a}\right) \end{aligned}

Proof:


Consider


\begin{aligned} I&=\int_{0}^{\infty} \cos x^{a} d x-i\int_{0}^{\infty} \sin x^{a} d x\\ &=\int_{0}^{\infty} e^{-i x^{a}} d x\\ &=\frac{1}{a} \int_{0}^{\infty} e^{-i x} x^{1 / a-1}\, d x & (x^{a} \to x)\\ &=\frac{1}{a} \frac{\Gamma\left(\frac{1}{a}\right)}{i^{1 / a}}\\ &= \frac{e^{-\frac{i \pi}{2 a}}}{a} \Gamma\left(\frac{1}{a}\right)\\ &=e^{-\frac{i \pi}{2 a}} \Gamma\left(1+\frac{1}{a}\right)\\ &=\left(\cos \left(\frac{\pi}{2 a}\right)-i \sin \left(\frac{\pi}{2 a}\right)\right) \Gamma\left(1+\frac{1}{a}\right) \qquad \blacksquare \end{aligned}


Equating Real and Imaginary parts We obtain the desired results

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