Calculate $\int_0^\infty\cos x^2dx$ using the fact that integral of complex functions on closed contour equals $0$

Question

We need to find value of: $$\int_0^\infty\cos x^2dx$$ Hint is to use $f(z)=\exp(iz^2)$. We know that $\cos x^2=f(x)+f(-x)$. So we can obtain it by some elementary operations on sum of integrals $\int_{-n}^0f(x)dx+\int_0^nf(x)dx$ and going with $n\to\infty$ afterwards. But how to add third integral to this sum in a way to close this curve?

Answer 1

Assuming that you have already proved that the integral is convergent (for instance by enforcing the substitution $x^2=t$ and exploiting Dirichlet's test) we have $$ I = \int_{0}^{+\infty}\cos(x^2)\,dx = \text{Re}\int_{0}^{+\infty} e^{ix^2}\,dx \tag{1}$$ and by setting $\eta=\frac{1+i}{\sqrt{2}}$ we have: $$ \int_{0}^{+\infty} e^{ix^2}\,dx = \eta\int_{0}^{\bar{\eta}\infty} e^{-z^2}\,dz \tag{2}$$ Now we may consider, for a large $M\in\mathbb{R}^+$, the contour $\gamma_M$ made by the line segment from $0$ to $\bar{\eta}M$, the arc of circle $\alpha_M$ from $\bar{\eta}M$ to $M$ and the line segment from $M$ to $0$. Since $e^{-z^2}$ is an entire function, $\oint_{\gamma_M}e^{-z^2}\,dz = 0$ and $$ \int_{0}^{M\bar{\eta}}e^{-z^2}\,dz = \int_{0}^{M}e^{-z^2}\,dz -\color{blue}{ \int_{\alpha_M}e^{-z^2}\,dz}.\tag{3} $$ It is simple to check that the blue term converges to zero as $M\to +\infty$ by the ML lemma, hence it follows that $$ \int_{0}^{+\infty}e^{ix^2}\,dx = \frac{1+i}{\sqrt{2}}\int_{0}^{+\infty}e^{-z^2}\,dz = \frac{1+i}{2}\sqrt{\frac{\pi}{2}}\tag{4} $$ and by considering the real part of both sides the Fresnel integral $$ \int_{0}^{+\infty}\cos(x^2)\,dx = \color{blue}{\sqrt{\frac{\pi}{8}}}\tag{5}$$ is evaluated.

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An equivalent approach is to notice that the Laplace transform gives $$ \int_{0}^{+\infty}\cos(x^2)\,dx = \int_{0}^{+\infty}\frac{\cos(t)}{2\sqrt{t}}\,dt \stackrel{\mathcal{L}}{=}\frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}}{1+s^2}\,ds = \frac{1}{2\sqrt{\pi}}\int_{-\infty}^{+\infty}\frac{u^2\,du}{1+u^4}\tag{6}$$ and the last integral is straightforward to evaluate through the residue theorem.