1. ## integrallll

$\int cos(x^2)dx$

2. ## Re: integrallll

Originally Posted by vernal
$\int cos(x^2)dx$

A primitive of $\cos x^{2}$ cannot be written in term of elementary functions... what You can try is to expand the $\cos x^{2}$ in McLaurin series and then integrate 'term by term'...

Kind regards

$\chi$ $\sigma$

3. ## Re: integrallll

I do not understand.

5. ## Re: integrallll

An integral of the type...

$C(x)=\int_{0}^{x} \cos u^{2}\ du$ (1)

... is known as 'Fresnel Integral' and some informations about them You can find in...

Fresnel Integrals -- from Wolfram MathWorld

The function C(*) as defined in (1) cannot be expressed in term of elementary functions but a way to pratically compute the (1) consists in using the McLaurin expansion...

$\cos u^{2}= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n)!}\ u^{4n}$ (2)

... and integrating 'term by term' obtaining...

$C(x)= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(4 n+1)\ (2 n)!}\ x^{4n+1}$ (3)

Kind regards

$\chi$ $\sigma$

chisigma

7. ## Re: integrallll

one of may friend said:

we have:
$e^{i\theta}=\cos(\theta)+i\sin(\theta)$

and
$\int_{-\infty}^{+\infty}e^{{-\alpha x^{2}}} dx=\sqrt{\frac{\pi}{\alpha}}$.

then we know:
$\int_{-\infty}^{+\infty}e^{{-i x^{2}}} dx=\int_{-\infty}^{+\infty} \cos(x^{2})-i \sin(x^{2})dx=\sqrt{\frac{\pi}{i}}$

$\sqrt{-i}=e^{-i \frac{\pi}{4}}\, \, \: \Rightarrow \: \sqrt{\frac{\pi}{i}}=\sqrt{-i \pi}=\sqrt{\cos(\frac{\pi}{4})\pi-i\sin(\frac{\pi}{4})\pi}=\frac{1}{2}\sqrt{\frac{\pi}{2}}-i\frac{1}{2}\sqrt{\frac{\pi}{2}}$

then

$\int_{0}^{\infty}\cos(x^{2})dx-i \int_{0}^{\infty}\sin(x^{2})dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}-i\frac{1}{2}\sqrt{\frac{\pi}{2}}$

$\int_{0}^{\infty}\cos(x^{2})dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$

Now my question is: Which is correct?
Both are probably true
What is the relationship between them?

8. ## Re: integrallll

Your friend didn't justify that $\int_{0}^{\infty} e^{-\alpha x^{2}} \ dx = \sqrt{\frac{\pi}{\alpha}}}$ is valid for imaginary $\alpha$.

$\int_{-\infty}^{\infty} e^{-ix^{2}} \ dx = \frac{1}{\sqrt{i}} \int^{\sqrt{i} \infty}_{-\sqrt{i} \infty} e^{-u^{2}} \ du$

or more approriately $\frac{1}{\sqrt{i}} \lim_{R \to \infty} \int^{\sqrt{i} R}_{-\sqrt{i} R} e^{-u^{2}} \ du$

$= \sqrt{\frac{\pi}{i}} \lim_{R \to \infty} \text{erf}( \sqrt{i} R)$ where erf(x) is the error function

We need to show that the limit evaluates to 1. The value of the error function at infinity is 1. But we have the error function at complex infinity. The limit can be evaluated by expanding the error function in a series near real infinity, substituting $\sqrt{i} R$ for x, and then taking the limit.

9. ## Re: integrallll

What RV says is true: the result obtained by vernal's friend is correct but he didn't previously demonstrate that both the integrals...

$\int_{0}^{\infty} \cos x^{2}\ dx\ ,\ \int_{0}^{\infty} \sin x^{2}\ dx$ (1)

... do exist. An example will clarify the concept: starting from the well known formula...

$\int_{0}^{\infty} e^{- \alpha x}\ dx = \frac{1}{\alpha}$ (2)

... I write...

$\int_{0}^{\infty} e^{-i x}\ dx = \int_{0}^{\infty} \cos x\ dx - i\ \int_{0}^{\infty} \sin x\ dx = \frac{1}{i}= -i$ (3)

... and from (3)...

$\int_{0}^{\infty} \cos x\ dx =0\ ,\ \int_{0}^{\infty} \sin x\ dx =1$ (4)

But the (4) are a nonsense because none of the integrals exist...

Kind regards

$\chi$ $\sigma$

10. ## Re: integrallll

I'm fairly certain that the fact that $\int_{-\infty}^{\infty} \cos x^{2} \ dx$ and $\int_{-\infty}^{\infty} \sin x^{2} \ dx$ converge doesn't by itself justify that $\int_{-\infty}^{\infty} e^{-ix^{2}} \ dx = \sqrt{\frac{\pi}{i}}$.

The justification is the uniqueness theorem from complex analysis.