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$$\begin{align}I&=\int_{0}^{\infty}{e^{-x^2}\cos(2x^2)\,dx}\\ I&\overset{u=2x^2}{=}\frac{\sqrt{2}}{4}\int_{0}^{\infty}e^{-\frac{u}{2}}\cos(u)\,\frac{du}{\sqrt{u}}\\I&=\Re\left(\frac{\sqrt{2}}{4}\int_{0}^{\infty}{e^{-u(\frac{1}{2}-i)}\,u^{\frac{1}{2}-1}\, du}\right)\end{align}$$

How can I continue here? It looks like the gamma function but I'm unsure how to eliminate the $(\frac{1}{2}-i)$ in this case. Usually it involves drawing a rectangular contour to justify a substitution but I don't think it can be applied at this stage (?)

AnthonyML
  • 967

3 Answers3

5

Let's denote $$I(a)=\int_0^\infty e^{-x^2}\cos(ax^2)\,dx=\frac12\Re\int_{-\infty}^\infty e^{-x^2(1-ia)}dx=\frac12\Re J(a)$$ Let's also consider $$J^2(a)=\int_{-\infty}^\infty e^{-x^2(1-ia)}dx\int_{-\infty}^\infty e^{-y^2(1-ia)}dy$$ Switching to the polar system of coordinates, $$J^2(a)=\int_0^\infty rdr\int_0^{2\pi}d\phi \,e^{-r^2(1-ia)}=\pi\int_0^\infty e^{-t(1-ia)}dt=\frac{\pi}{1-ia}=\frac{\pi}{\sqrt{1+a^2}}e^{i\arctan a}$$ Hence, $$J(a)=\frac{\sqrt\pi}{\sqrt[4]{1+a^2}}e^{\frac{i}{2}\arctan a}$$ $$I(a)=\frac12\Re J(a)=\frac{\sqrt\pi}{2\sqrt[4]{1+a^2}}\cos\big(\frac{1}{2}\arctan a\big)=\frac{\sqrt\pi}{2\sqrt[4]{1+a^2}}\sqrt{\frac{1+\cos(\arctan a)}{2}}$$ Using also $\,\displaystyle1+\tan^2a=\frac{1}{\cos^2a}$ $$\boxed{\,\,I(a)=\int_0^\infty e^{-x^2}\cos(ax^2)\,dx=\frac{\sqrt\pi}{2\sqrt2}\frac{\sqrt{1+\sqrt{1+a^2}}}{\sqrt{1+a^2}}\,\,}$$ For $a=2$ $$I(2)=\frac{\sqrt\pi\sqrt{1+\sqrt5}}{2\sqrt{10}}$$

Svyatoslav
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3

You're almost there. Simply put $v=u(\frac12-i),u=v(\frac{2+4i}5)$ into your $u$ integral. You should get $\Gamma(\frac12)=\sqrt\pi$ times a constant, whose real part is then the required answer.

One barrier is you need to denest $\sqrt{2+4i}$. For this purpose you may use the formula given here. Thus

$\sqrt{2+4i}=\dfrac{2\sqrt5+2+4i}{\sqrt{2(2\sqrt5+2)}}.$

Oscar Lanzi
  • 39,403
  • Great answer, could you quickly explain why writing $\sqrt{2+4i}$ in this form is preferable to writing $\sqrt{2\sqrt{5}e^{i \arctan(2)}}$? – AnthonyML Mar 07 '23 at 21:30
  • You can use algebraic functions to obtain square roots of complex numbers, including the formula I reference. Then you get the roots without having to invoke trig or inverse trigonometric functions. Square roots (or repeating these like fourth roots) are the only roots of complex numbers where you can generally do this. – Oscar Lanzi Mar 07 '23 at 21:36
1

$$I(a)=\int_0^\infty e^{-x^2}\cos(ax^2)\,dx=\Re\left(\int_{0}^\infty e^{-x^2(1-ia)}dx\right)$$ $$\int_{0}^\infty e^{-k x^2}\,dx=\frac{\sqrt{\pi }}{2 \sqrt{k}}\qquad\text{ if }\qquad\Re(k)>0$$ $$k=1-i a \implies \frac{1}{\sqrt{k}}=\frac{e^{-\frac{1}{2} i \arg (1-i a)}}{\sqrt[4]{a^2+1}}\implies$$ $$\Re\left(\frac{1}{\sqrt{k}}\right) =\frac{\cos \left(\frac{1}{2} \arg (1-i a)\right)}{\sqrt[4]{a^2+1}}$$

Just play with the complex numbers to simplify the result.

Claude Leibovici
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