Complex analysis: showing that contour integration of function cos and sin are equal

Question

I need to show that $\int x\cos(x^4) = \int x\sin(x^4) = (1/4)\sqrt{ \pi/2}$. The bounds of both integrals are $0$ to $\infty$.

Should I first let $f(z) = ze^{iz^4}$?

Answer 1

Enforcing the substitution $x\to \sqrt{x}$ reveals

$$\int_0^\infty xe^{ix^4}\,dx=\frac12 \int_0^\infty e^{ix^2}\,dx \tag 1$$

Since $e^{iz^2}$ is entire, Cauchy's Integral Theorem guarantees that $\oint_C e^{iz^2}\,dz=0$ for any rectifiable curve $C$. Letting $C$ be the "pie wedge" contour with edges from $0$ to $R$ and from $Re^{i\pi/4$}$ to $0$, we find that

$$\begin{align} \int_0^\infty e^{ix^2}\,dx&=\lim_{R\to \infty}\int_0^R e^{ix^2}\,dx\\\\ &=\lim_{R\to \infty}\int_0^R e^{i(e^{i\pi/4}y)^2}\, e^{i\pi/4}\,dy-\lim_{R\to \infty}\int_0^{\pi/4}e^{i(Re^{i\phi})^2}\,iRe^{i\phi}\,d\phi\\\\ &=e^{i\pi/4}\int_0^\infty e^{-y^2}\,dy\\\\ &=e^{i\pi/4}\frac{\sqrt{\pi}}{2}\tag 2 \end{align}$$

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Substituting $(2)$ into $(1)$ and equating real and imaginary parts yields the coveted results

$$\int_0^\infty x\cos(x^4)\,dx=\frac14\sqrt{\frac{\pi}{2}}$$

and

$$\int_0^\infty x\sin(x^4)\,dx=\frac14\sqrt{\frac{\pi}{2}}$$