# Some way to integrate $\sin(x^2)$? [duplicate]

Because the straight forward approach involves Fresnel integrals I thought about a different approach of taking the imaginary part of $\int_{-\infty}^{\infty}\exp{(ix^2)} \, dx$ but have no idea how to continue.

Knowing that $\int_{-\infty}^{\infty}\exp{(-x^2)} \, dx = \sqrt \pi$ one can feel that this value has to be somehow "divided" over the entire complex plane, so that every quadrant gets some part of $\sqrt \pi$. This would give the desired result of $\sqrt \frac{\pi}{2} - i \sqrt \frac{\pi}{2}$

Is there a way this could be proved or justified?

• Instead of "knowing" $\int_{-\infty}^{\infty}\exp(-x^2)\,dx=\sqrt{\pi}$ if you knew how this formula was derived, you'd be able to modify the argument to get what you want. So, look for a proof of the integral you know. – Gerry Myerson Dec 5 '12 at 0:03
• The question that this is marked as a duplicate specifically asks for real methods. This question asks how to proceed on a particular complex integration approach. – robjohn Jul 21 '18 at 3:28

Check out this answer for a real method to evaluate this integral.

First note that $$\int_0^\infty e^{-x^2}\,\mathrm{d}x=\frac{\sqrt\pi}2\tag{1}$$

The integral $$\int_\gamma e^{-z^2}\,\mathrm{d}z=0\tag{2}$$ over the curve $\gamma$ consisting of the line from $0$ to $R$ then counterclockwise along the circular arc from $R$ to $Re^{i\pi/4}$ then back along the line from $Re^{i\pi/4}$ to $0$ must be $0$ since $e^{-z^2}$ has no singularities inside $\gamma$.

Next note that $\int_\gamma e^{-z^2}\,\mathrm{d}z$ along the line from $0$ to $R$ as $R\to\infty$ tends to $(1)$.

Next note that $\int_\gamma e^{-z^2}\,\mathrm{d}z$ along the arc of the circle of radius $R$ from $R$ to $Re^{i\pi/4}$ as $R\to\infty$ goes to zero.

Finally, note that $\int_\gamma e^{-z^2}\,\mathrm{d}z$ along the line from $Re^{i\pi/4}$ to $0$ is $$-e^{i\pi/4}\int_0^\infty e^{-ix^2}\,\mathrm{d}x\tag{3}$$ Therefore, $(1)$ plus $(3)$ is $0$, so we get $$\int_0^\infty e^{-ix^2}\,\mathrm{d}x=\frac{1-i}{\sqrt2}\frac{\sqrt\pi}2\tag{4}$$ Taking the imaginary part of $(4)$ yields that $$\int_0^\infty\sin(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{5}$$ or $$\int_{-\infty}^\infty\sin(x^2)\,\mathrm{d}x=\sqrt{\frac\pi2}\tag{6}$$

• What was the motivation behind taking the contour as in you answer? Why not a different contour like a semi-circle? I took the semi-circle and the integral turned out to be difficult for me. – Apurv Jun 8 '16 at 9:08
• The point is to relate the known integral $\int_0^\infty e^{-x^2}\,\mathrm{d}x$ to the unknown integral $\int_0^\infty e^{ix^2}\,\mathrm{d}x$ while making sure that the integral near $\infty$ vanishes. If we take the whole semi-circle, the integrand blows up along parts of the semi-circle. – robjohn Jun 8 '16 at 10:48
• @robjohn do you mind expanding on the part along the diagonal? I am trying use a parameterization for $z=xe^{i \pi/4}$ for x from 0 to R and I am not getting the correct result. – MathIsHard Jul 20 '18 at 19:58
• @MathIsHard: $$\int_\infty^0e^{-\left(xe^{i\pi/4}\right)^2}\,\mathrm{d}xe^{i\pi/4} =-e^{i\pi/4}\int_0^\infty e^{-ix^2}\,\mathrm{d}x$$ and as described above, $$\int_\infty^0e^{-\left(xe^{i\pi/4}\right)^2}\,\mathrm{d}xe^{i\pi/4} =-\int_0^\infty e^{-x^2}\,\mathrm{d}x$$ Equating the right sides, we get $$\int_0^\infty e^{-ix^2}\,\mathrm{d}x = e^{-i\pi/4}\int_0^\infty e^{-x^2}\,\mathrm{d}x$$ – robjohn Jul 21 '18 at 6:16
• @robjohn thank you so much! I really appreciate it. :) – MathIsHard Jul 21 '18 at 15:54

I think robjohn's answer is excellent, but I also think it would be helpful to provide the details explaining why the integral along the arc vanishes as $R \to \infty$.

For the path along the arc we may choose $\gamma(t) = R e^{i\frac{\pi}{4}t}$ for $0 \leq t \leq 1$. We will show $$\lim_{R \to \infty} \bigg| \int_\gamma e^{-z^2} dz \bigg| = 0.$$ For fixed $R$, we have $$\bigg| \int_\gamma e^{-z^2} dz \bigg| \leq \int_\gamma \bigg| e^{-z^2} \bigg| d|z| = \int_0^1 \bigg| e^{-[\gamma(t)]^2} \bigg| \big| \gamma^\prime(t) \big| dt = R \frac{\pi}{4} \int_0^1 \bigg| e^{-R^2 e^{i \frac{\pi}{2} t}} \bigg| dt.$$ Decomposing the exponential in the numerator, this becomes $$R \frac{\pi}{4} \int_0^1 e^{-R^2 \cos(\frac{\pi}{2}t)} dt \leq R \frac{\pi}{4} \int_0^1 e^{-R^2(1-t)} dt = R \frac{\pi}{4} \int_0^1 e^{-R^2 u} du = \frac{\pi}{4} \frac{1}{R} (1 - e^{-R^2}).$$ Hence, $$\lim_{R \to \infty} \bigg| \int_\gamma e^{-z^2} dz \bigg| \leq 0,$$ as desired.

• How does one get the estimate $\cos\left(\frac\pi2 t\right)\geq 1-t$? Drawing the things it's obvious. – leo Jul 23 '15 at 18:08
• Oh well it's concavity. – leo Jul 23 '15 at 18:09
• I am curious if the parameterization of $z=Re^{i \theta}$ would work for $\theta$ from 0 to $\pi/4$? If so, what made you decide to pick the parameterization that you did? Thanks. – MathIsHard Jul 20 '18 at 19:04

Starting from $\int_{-\infty}^\infty \exp(-x^2)\ dx = \sqrt{\pi}$, use a change of variables to get $\int_{-\infty}^\infty \exp(-ax^2)\ dx = \sqrt{\pi/a}$ for $a > 0$. Now both sides are analytic in $a$ for $\text{Re}(a) > 0$ (using a branch of the square root that is analytic in the right half plane), so the equation should still be true for $\text{Re}(a) > 0$. The next part is tricky to justify: you want to take the limit as $a \to -i$. Assuming this is valid, you have

$$\int_{-\infty}^\infty \exp(ix^2)\ dx = \sqrt{\frac{\pi}{-i}}= (1+i) \sqrt{\frac{\pi}{2}}$$

so that $\int_{-\infty}^\infty \cos(x^2)\ dx = \int_{-\infty}^\infty \sin(x^2)\ dx = \sqrt{\pi/2}$.

Hint: you should integrate $\exp(-x^2)$ along the Fresnel contour; the integral along the real axis you know, the integral along the diagonal you want to find, the arc at infinity vanishes.