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So, I need to prove the identity

$$\int_{-\infty}^\infty \cos t^2 dt = \int_{-\infty}^\infty \sin t^2 dt = \sqrt{\frac{\pi}{2}}$$

and as a hint I have the Gaussian integral

$$\int_{-\infty}^\infty e^{-xt^2} dt = \sqrt{\frac{\pi}{x}} \;\;\;\forall x>0.$$

I suspect I have to take the real/imaginary part of $e^{-t^2}$ at some point, but I can't quite figure how. I.e., $\int e^z dz = e^z$ gives me nothing. So, how do I do it?

Minethlos
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3 Answers3

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For a nice real analytic derivation, we can notice that: $$I^2=\left(\int_{0}^{+\infty}\cos(t^2)\,dt\right)^2 = \frac{1}{2}\iint_{[0,+\infty)^2}\cos(s^2-t^2)\,ds\,dt = \frac{1}{4}\int_{0}^{+\infty}\int_{-v}^{v}\cos(uv)\,du\,dv$$ so: $$ I^2 = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin v^2}{v}\,dv =\frac{1}{4}\int_{0}^{+\infty}\frac{\sin z}{z}\,dz=\frac{\pi}{8}.$$ Some care is needed in proving that $\iint_{[0,+\infty)^2}\cos(s^2+t^2)\,ds\,dt$ vanishes and the manipulations are legit, since $\frac{\sin z}{z}$ is a Riemann integrable function but it does not belong to $L^1(\mathbb{R})$. Beside that, this proof just mimics the proof that $\int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}$ through the Fubini-Tonelli's theorem.

Jack D'Aurizio
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  • As I just commented to the question, I have a real-variable approach. Do you have a nice real-variable method to evaluate $\int_0^\infty\frac{\sin(x)}{x}\mathrm{d}x$? – robjohn Jan 10 '15 at 22:27
  • @robjohn: I usually prove it with the (inverse) Laplace transform or through the identity $$\sum_{n=-\infty}^{+\infty}\frac{(-1)^n}{x+n\pi}=\frac{1}{\sin x}.$$ – Jack D'Aurizio Jan 10 '15 at 22:32
  • @robjohn: in fact, we can deal with convergence issues by introducing an exponential smoothing factor $e^{-\alpha x}$ and letting $\alpha\to 0$. This gives that our approaches are essentially the same. – Jack D'Aurizio Jan 10 '15 at 22:46
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    @JackD'Aurizio: that is most likely so since my approach also mimics the classic proof of $\int_{-\infty}^\infty e^{-\pi x^2},\mathrm{d}x=1$ – robjohn Jan 10 '15 at 22:55
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HINT:

$\Im e^{iz} = \sin(z) \implies \sin(z^2) = \Im e^{iz^2}$

Amad27
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  • Yes, I know Euler's identity. But I don't see any straightforward way to use it here. Do you mean there is simpler solution than here? – Minethlos Jan 10 '15 at 21:27
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Substitute $x=i$ in your hint and simplify the resulting complex number and then you need to consider real and imaginary parts of the integral. See here for a justification which it depends on your background.

Community
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Mhenni Benghorbal
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