Fresnel integral

The Fresnel integrals read

(1)   \begin{align*} {\cal S} \,=\, \int_0^{\infty}dt\, \sin t^2 \\ {\cal C} \,=\, \int_0^{\infty}dt\, \cos t^2 \end{align*}

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Consider the following integral

(2)   \begin{align*} {\sqrt{\pi} \over 2} \,=\, \int_0^{\infty}dz\,e^{-z^2} \,&=\, -\int_{\gamma_3}dz\,e^{-z^2} \\ \,&=\, e^{i\pi/4} \int_0^{\infty}dt\,e^{-it^2} \\ \,&=\, {1+i \over \sqrt{2}} \int_0^{\infty}dt\,\big(\cos t^2 - i\sin t^2\big) \\ \,&=\, {1 \over \sqrt{2}} \int_0^{\infty}dt\,\Big( \big(\cos t^2 + \sin t^2\big) +i \big(\cos t^2 - \sin t^2\big)\Big) \\ \,&=\, {1 \over \sqrt{2}} \Big( \big({\cal C} + {\cal S}\big) +i \big({\cal C} - {\cal S}\big)\Big) \end{align*}

From this we have

(3)   \begin{align*} {\cal C} + {\cal S} \,&=\, {\sqrt{\pi} \over \sqrt{2}} \\ {\cal C} - {\cal S} \,&=\, 0 \end{align*}

which gives

(4)   \begin{align*} {\cal C} \,=\, {\cal S} \,=\, {\sqrt{\pi} \over \sqrt{8}} \end{align*}

Equation (2) also gives

(5)   \begin{align*}   \int_{-\infty}^{\infty}dt\,e^{-it^2} \,=\, e^{-i\pi/4} \int_{-\infty}^{\infty}dz\,e^{-z^2}  \,=\, \sqrt{\pi} e^{-i\pi/4} \end{align*}

We give an alternative derivation in the following. Let us start with

(6)   \begin{align*} G(x) \,=\, \bigg(\int_0^{x}dt\, e^{it^2}\bigg)^2 + i\int_0^1 dt\, {e^{ix^2(t^2+1)} \over t^2 + 1} \end{align*}

Differentiating G(x) with respect to x

(7)   \begin{align*} {dG(x) \over dx} \,&=\, 2\bigg(\int_0^{x}dt\, e^{it^2}\bigg)e^{ix^2}  + i\int_0^1 dt\, {i2x(t^2+1) e^{ix^2(t^2+1)} \over t^2 + 1} \\ \,&=\, 2\bigg(\int_0^{x}dt\, e^{it^2}\bigg)e^{ix^2}  -2x \int_0^1 dt\, e^{ix^2(t^2+1)} \\ \,&=\, 2\bigg(\int_0^{x}dt\, e^{it^2}\bigg)e^{ix^2}  -2e^{ix^2}\int_0^x du\, e^{iu^2} \,=\, 0 \end{align*}

where we chnage the variable, u=xt.
Therefore we have

(8)   \begin{align*} G(x) \,\equiv\, G(0) \,=\, i\int_0^1 dt {1 \over t^2 + 1} \,=\, i{\pi \over 4} \end{align*}

Since in the limit x\to \infty

(9)   \begin{align*} \lim_{x\to\infty}\int_0^1 dt\, {e^{ix^2(t^2+1)} \over t^2 + 1} \,=\, 0 \end{align*}

we have

(10)   \begin{align*} G(\infty) \,=\, \bigg(\int_0^{\infty}dt\, e^{it^2}\bigg)^2 \,=\, \bigg({\cal C} + i {\cal S}\bigg)^2  \end{align*}

Solving the equation

(11)   \begin{align*} i{\pi \over 4} \,=\, {\cal C}^2 - {\cal S}^2 + 2i\, {\cal C}{\cal S} \end{align*}

gives (4).

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