# Rational Fresnel -type integral.

I have a question about an integral that looks like a great candidate for residues.

$\displaystyle \int_{0}^{\infty}\frac{\cos(x^{2})}{x^{4}+1}dx-\int_{0}^{\infty}\frac{\sin(x^{2})}{x^{4}+1}dx=\frac{\pi\sqrt(2)}{4e}$.

My difficulty arises in knowing how to come up with the proper contour for something like this.

Does anyone know of a good method for evaluating the above integral using residue theory, or even real methods?.

Take the cosine one.

I wrote it out as $\displaystyle\int_{C}\frac{e^{iz^{2}}}{z^{4}+1}dz$.

Where C is the semicircular contour in the upper half plane.

The zeroes of the denominator are $\displaystyle e^{\frac{\pi i}{4}}, \;\ e^{\frac{3\pi i}{4}}, \;\ e^{\frac{5\pi i}{4}}, \;\ e^{\frac{7\pi i}{4}}$. Of which the first two lie in the upper half plane.

The residue at $\displaystyle e^{\frac{\pi i}{4}}$ is $\frac{-\sqrt{2}}{8e}-\frac{\sqrt{2}i}{8e}$

The residue at $\displaystyle e^{\frac{3\pi i}{4}}$ is $\frac{e\sqrt{2}}{8}-\frac{e\sqrt{2}i}{8}$

Summing them: $\displaystyle 2\pi i\left(\frac{-\sqrt{2}}{8e}-\frac{\sqrt{2}i}{8e}+\frac{e\sqrt{2}}{8}-\frac{e\sqrt{2}i}{8}\right)$

$\displaystyle =\frac{(e^{2}+1)\pi\sqrt{2}}{4e}+\frac{(e^{2}-1)\pi\sqrt{2}}{4e}\cdot i$

But, this is not the correct result... numerically. No doubt, it is not that straightforward because I have chosen an incorrect contour. Perhaps due to the Fresnel-type term in the numerator. Can anyone lend any insight on this one?.

Thanks very much.

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## 1 Answer

Try the quarter-circle contour that goes out on the positive real axis, turns counter-clockwise, and then comes down the positive imaginary axis. The motivation is to respect symmetry of the integrand.

The only residue enclosed by this contour, which I'll call $C$, is at $e^{i\pi/4}$, which you've already computed as $-\frac{\sqrt{2}}{8e}(1+i)$. Then we have $$\frac{\pi\sqrt{2}}{4e}(1-i)=\int_C dz\,\frac{e^{iz^2}}{z^4+1}=\int_1+\int_2+\int_3$$ where $\int_1$, $\int_2$, $\int_3$ refer to the integrals over the three pieces of the contour. $1$ is going out on the real axis, $2$ is the circular arc going CCW by $\pi/2$ and $3$ is going down the imaginary axis. The circular piece, $2$, contributes $0$ in the limit of the contour getting really big, so we won't bother with it.

Now the interesting part is $3$, where we parametrize by $z=it$, $t=\infty\ldots 0$ and compute $$\int_3=\int_3 dz\,\frac{e^{iz^2}}{z^4+1}=-\int_0^\infty i dt\,\frac{e^{-it^2}}{t^4+1}.$$ This is $-i$ times the complex conjugate of $\int_1$. Hence the residue theorem says $$\frac{\pi\sqrt{2}}{4e}(1-i)=\int_1-i\left(\int_1\right)^*.$$ The integral in question is really $I={\rm Re}(\int_1)$. Take the real and imaginary parts of the equation above and solve for $I$ to find $$I=\frac{\pi\sqrt{2}}{4e}.$$

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Sure, Jonathan, more details would be nice if you feel like it. I am always interested in learning something new about integrating with residues. The standard Fresnel integral is rather famous. It must be something like the way it is done with the quarter circle then?. Thanks for your fast response. – Cody Dec 9 '12 at 16:55
@Cody, I've added some details. I hope it helps. – Jonathan Dec 9 '12 at 17:20
It sure does. Thanks a lot. – Cody Dec 9 '12 at 19:20