MHB How to Calculate $\int_0 ^{\infty} \sin t^2$ Using Complex Analysis?

[Jakob1]

Hello.

I'm having trouble calculating $\int_0 ^{\infty} \sin t^2$ using the fact that $\int _{\partial Tr} e^{-z^2} dz = 0$, where $Tr = conv (\{ 0, r, r + ir \})$ (a triangle).

I'm aware that I need to somehow transform $e^{-z^2}$ to get $\sin t^2, \ \cos t^2$ but I don't know how to do that.

It doesn't help if I write $e^z = e^x (\cos y + i \sin y)$, because then $e^{-z^2} = e^{y^2 - x^2} (\cos 2xy - i \sin 2xy)$, so there's no way I can get a square of $t$ that way.

How can one relate $e^z$ to $\sin t, \ \sin t^2$ differently from what is written above? I also don't see how I can use the triangle to compute this integral.

Could you help me with that?

 

[alyafey22]

$$e^{-iz^2} = \cos(z^2)-i\sin(z^2)$$

On the interval $[0,r]$ we have

$$\int^r_0 e^{-iz^2} \,dz = \int^r_0 e^{-it^2}\,dt = \int^r_0 \cos(t^2)\,dt-i\int^r_0 \sin(t^2)\,dt$$

 

[Jakob1]

Thank you.

So I know that on the intervals $[0,r]$ the integrals $$\int^r_0 e^{-iz^2} \,dz = \int^r_0 e^{-it^2}\,dt = \int^r_0 \cos(t^2)\,dt-i\int^r_0 \sin(t^2)\,dt$$ are equal.

But on this interval $[r, r+ir]$ we have $z= (r + ir)y$ so $dz=(1+i)dy$ and $z^2 = (ry+iry)^2$, so here we get:

$$ \int ^{r+ir}_r e^{-(ry+iry)^2}idy$$. This doesn't seem to be easy to compute or transform into $\sin, \ \cos$...

we also have that $$\int^{r+ir}_r e^{-iz^2} \,dz = \int^{r+ir}_r \cos z^2 dz+ i \int^{r+ir}_r \sin z^2 dz$$

 

[alyafey22]

Thank you.

So I know that on the intervals $[0,r]$ the integrals $$\int^r_0 e^{-iz^2} \,dz = \int^r_0 e^{-it^2}\,dt = \int^r_0 \cos(t^2)\,dt-i\int^r_0 \sin(t^2)\,dt$$ are equal.

But on this interval $[r, r+ir]$ we have $z= r + iry$ so $dz=idy$ and $z^2 = (r+iy)^2$, so here we get:

$$ \int ^{r+ir}_r e^{-(r+iry)^2}idy$$. This doesn't seem to be easy to compute or transform into $\sin, \ \cos$...

we also have that $$\int^{r+ir}_r e^{-iz^2} \,dz = \int^{r+ir}_r \cos z^2 dz+ i \int^{r+ir}_r \sin z^2 dz$$

Sorry , I just realized that in your question we have to use the function $f(z) = e^{-z^2}$.

So have you done the parametrizations ? what do you get ?

 

[Jakob1]

If we go this way: $$0 \rightarrow r \rightarrow r+ir$$

we get:

for the segment $$[0, \ r] : \ z=x$$, so $dz = dx$ and $z^2 = x^2$

$$ [r, \ r+ir] : \ z= r + iry$$, so $dz=idy$ and $z^2 = (r+iry)^2$.

So the integral has this form: $$\int_0^r e^{-x^2} dx + i \int_r^{r+ir} e^{-(r+iry)^2} dy$$.

I don't know what to do next. Could you help?

 

[alyafey22]

If we go this way: $$0 \rightarrow r \rightarrow r+ir$$

we get:

for the segment $$[0, \ r] : \ z=x$$, so $dz = dx$ and $z^2 = x^2$

$$ [r, \ r+ir] : \ z= r + iry$$, so $dz=idy$ and $z^2 = (r+iry)^2$.

So the integral has this form: $$\int_0^r e^{-x^2} dx + i \int_r^{r+ir} e^{-(r+iry)^2} dy$$.

I don't know what to do next. Could you help?

You have done the parametrization the wrong way. It should be $z = (1+i)y$ . The integral then becomes

$$ (1+i) \int_0^{r} e^{-((1+i)y)^2} dy$$

 

[Jakob1]

I see.

You've determined the equation of the line on which lie $$0$$ and $$r+ir$$.However, I still do not see how to use it to derive $ \sin t^2$ from this integral(s).

 

[alyafey22]

Sine the function $f$ is analytic the integral along the closed path (triangle) must be zero.

$$\int^r_0 e^{-(1+i)^2y^2}\,dy = \int^r_0 e^{-2iy^2}\,dy$$

which is more or less the integral we want to evaluate. It remains to find the integral along the vertical line of the triangle. You need to re-evaluate it because you have done it wrong.

 

[Jakob1]

I've noticed my mistake and I've just edited my post. Could you tell me if it is all right now?

On this interval $[r, r+ir]$ we have $z= (r + ir)y$ so $dz=(1+i)dy$ and $z^2 = (ry+iry)^2$, so here we get:

$$ \int ^{r+ir}_r e^{-(ry+iry)^2}idy = \int ^{r+ir}_r e^{-2r^2y^2} i dy$$Could we use this $$e^{-iz^2} = \cos(z^2)-i\sin(z^2)$$ now?

 

[alyafey22]

I've noticed my mistake and I've just edited my post. Could you tell me if it is all right now?

On this interval $[r, r+ir]$ we have $z= (r + ir)y$ so $dz=(1+i)dy$ and $z^2 = (ry+iry)^2$, so here we get

To make sure that you are doing it the right way substitute the limits of integration. You used the lower limit as $r$ but when you substitue in $z=(r+ir)y$ you get $z=(r+ir)r$ which is not the starting point of the vertical line. The same goes for the upper limit.

 

[Jakob1]

I'm sorry for the confusion. Let me try one more time.

The equation of the line passing through $r$ and $r+ir$ is $x=r$, so $z=r+iy$, (because $ z=x+iy$), so $dz=idy$ and $ z^2 = (r+iy)^2=r^2-y^2+2iry$, so the integral over the vertical segment is $$\int_0^r (e^{y^2-r^2-2iry})dy$$.

Is it correct this time?

 

[alyafey22]

I'm sorry for the confusion. Let me try one more time.

The equation of the line passing through $r$ and $r+ir$ is $x=r$, so $z=r+iy$, (because $ z=x+iy$), so $dz=idy$ and $ z^2 = (r+iy)^2=r^2-y^2+2iry$, so the integral over the vertical segment is $$\int_0^r (e^{y^2-r^2-2iry})dy$$.

Is it correct this time?

Yes , you just forgot the i from differentiation. Now , you need to prove that this integral vanishes as $r \to \infty$.

 

[Jakob1]

$$\int_0^r e^{y^2-r^2-2iry} dy = e^{-r^2} \int_0^r e^{y^2-2iry} dy $$

I thought I could estimate it somehow or use the error function (this integral is equal to $$\frac{\pi}{2}(erf(x)+erf(1+i)r)$$), but I haven't come up with anything so far.

 

[alyafey22]

$$\int_0^r e^{y^2-r^2-2iry} dy = e^{-r^2} \int_0^r e^{y^2-2iry} dy $$

I thought I could estimate it somehow or use the error function (this integral is equal to $$\frac{\pi}{2}(erf(x)+erf(1+i)r)$$), but I haven't come up with anything so far.

Hint 1: get rid of the exponential containing $i$.
Hint 2: use L'Hospital rule.

 

[Prove It]

Would this question be easier to do over a different contour? Like a circle for example?

 

[alyafey22]

Would this question be easier to do over a different contour? Like a circle for example?

Do you mean a semi-circle ?

I know that it can also be found using a sector (Fresnel integral) but the idea is the same.

 

[Jakob1]

Should I use de l'Hospital's rule to the function I'm integrating? If so, I get $$\frac{-2re^{y^2-r^2}}{-2( \sin 2yr - i \cos 2yr)}$$. Did I misunderstand you?
What about $\sin t^2$?

 

[alyafey22]

First you need to use the property $|e^{ix}|\leq 1$ , then apply the differentiation.

 

[Jakob1]

I need to prove that this integral $$\int_{0}^{r} \frac{e^{y^2 - r^2}}{e^{2ir}}\,d y$$ vanishes as $r$ approaches infinity.

$$| e^{y^2 - r^2}e^{-2ir} | \le | e^{y^2-r^2} | $$, because $$|e^{-2ir}| \le 1$$.

But now I don't see how I can apply the de l"Hospital rule, because there is no denominator.

 

[alyafey22]

I need to prove that this integral $$\int_{0}^{r} \frac{e^{y^2 - r^2}}{e^{2ir}}\,d y$$ vanishes as $r$ approaches infinity.

$$| e^{y^2 - r^2}e^{-2ir} | \le | e^{y^2-r^2} | $$, because $$|e^{-2ir}| \le 1$$.

But now I don't see how I can apply the de l"Hospital rule, because there is no denominator.

Good. Now you get

$$| \int^r_0 e^{y^2 - r^2}e^{-2ir} \, dy | \le \frac{\int^r_0 e^{y^2}\,dy}{e^{r^2}}$$

$$\lim_{r \to \infty} \frac{\int^r_0 e^{y^2}\,dy}{e^{r^2}} = \lim_{r \to \infty}\frac{e^{r^2}}{2r e^{r^2}} \to 0$$

 

[Jakob1]

Ok, I see that now.

Now there remain only two integrals:

$$\int_0^{\infty} e^{-x^2} dx = (1+i) \int_{0}^{\infty} e^{-2iy^2} \,d y = (1+i) \int_{0}^{\infty} (\cos 2y^2 - i \sin 2y^2) \,d y$$.

$$= \int_{0}^{\infty} \cos 2x^2 \,d x - i \int_{0}^{\infty} \sin 2x^2 \,d x +i \int_{0}^{\infty} \cos 2x^2 \,d x + \int_{0}^{\infty} \sin 2x^2 \,d x$$

So $$\int_{0}^{\infty} \cos 2x^2 \,d x - \int_{0}^{\infty} \sin 2x^2 \,d x =0$$

and $$\int_{0}^{\infty} \cos 2x^2 \,d x + \int_{0}^{\infty} \sin 2x^2 \,d x = \int_0^{\infty} e^{-x^2} dx $$.

Is that correct?

If it is, then $$\int_{0}^{\infty} \cos 2x^2 \,d x = \int_{0}^{\infty} \sin 2x^2 \,d x = \frac{1}{2} \int_0^{\infty} e^{-x^2} dx $$

Ok, I checked my result on wolfram :)

Thank you very, very much for all your help!

 

[alyafey22]

Ok, I checked my result on wolfram :)

Thank you very, very much for all your help!

very nice. I hope every think is clear , now.