yip said:
For that integral, i think u have to use the partial fraction theorem that involves derivatives, as there is trigonometric terms in the primitive of that function. Here is how I did the second integral, since u asked, the trick is to apply euler's theorem:
e^{-ix^{2}}=cosx^{2}-isinx^{2}
\int_{0}^{\infty}e^{-ix^{2}}dx=\int_{0}^{\infty}cosx^{2}dx-i\int_{0}^{\infty}sinx^{2}dx
It is known that \int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}
\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}
\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}
\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)
\int_{0}^{\infty}sinx^{2}dx=I\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}
This is oversimplifying the proof. The usual trick where gaussian integral is calculated as a square root of the two dimensional gaussian integral, gives the equation
<br />
\int\limits_{-\infty}^{\infty} e^{-(A+iB)x^2}dx = \sqrt{\frac{\pi}{A+iB}}<br />
in which A and B are real, only when A > 0. (The square root is the one that is continuous on set \{z\in\mathbb{C}\;|\; \textrm{Re}(z)>0\}, and that agrees with the positive square root on real axis.) If A=0, then the substitution r=\infty does not converge in the two dimensional integral, and thus the result is not trivially the same.
However, the Leibniz's rule of alternating series can be used to show that integrals
<br />
\int\limits_{0}^{\infty}\cos(x^2)dx<br />
and
<br />
\int\limits_{0}^{\infty}\sin(x^2)dx<br />
converge towards something, and thus the integral
<br />
\int\limits_{-\infty}^{\infty} e^{ix^2}dx<br />
also exists. If we know that the integral is a continuous function of the coefficent in the exponent, then this is all done, but I never bothered actually try to figure out how the continuity could be proven. If it can be done nicely, feel free to mention how.
