Can integrals be solved by rotation of axis?

[cragar]

You know how some integrals can't be solved. But is it possible to solve some of these integrals by rotation of axis or does it just make it easier.
Like in this link about half way down the page they found an integral equivalent to the Riemann zeta function. But then they had to do rotation by axis to do the integral. Could they have done it without rotation by axis.
http://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html

 

[Samuelb88]

No, they couldn't have solved the integral you're referring to without rotating the region of integration by $$\frac{\pi}{4}$$. This is because the original limits of integration are improper at the upper limits. In fact, the rotation you're referring to is actually a transformation $$x=x(u,v)$$, and $$y=y(u,v)$$. Transformations can be used to change untamable regions of integration into tamable ones.

A famous example of a seemingly unsolvable integral is $$\int_0^{\infty} e^{-x^2} dx$$. But in fact it can be solved. This integral is called the Gaussian integral, should you be interested in further reading.

 

[Quinzio]

A famous example of a seemingly unsolvable integral is $$\int_0^{\infty} e^{-x^2} dx$$. But in fact it can be solved. This integral is called the Gaussian integral, should you be interested in further reading.

Which is solved by changing the variables and using polar coordinates, if I'm not wrong.
Very elegant.

 

[cragar]

But its weird in the Gaussian integral they square the whole thing first, then change it to polar and then square root the answer. Is that even allowed to square it first? it only works on some integrals.

 

[Samuelb88]

Of course we are allowed to square an integral. Why would we not be able to?

 

[cragar]

Because it leads to different answers. I can't just square any integral then integrate it and then take the square root to get my answer.

 

[Samuelb88]

I think you can. Suppose $$f(x)$$ is a continuous function on [a,b]. Then $$\int_a^b f(x) dx = F(b)-F(a)$$. Now let $$I = \int_a^b f(x) dx = F(b)-F(a)$$. Then $$I^2 = (\int_a^b f(x) dx)(\int_a^b f(x) dx)= (F(b)-F(a))(F(b)-F(a)) = (F(b)-F(a))^2$$. Hence $$I = F(b)-F(a)$$. As an application, I will cite:

Let $$f(x)=x^2$$ and let $$I = \int_1^2 x^2 dx$$. Then $$I^2 = (\int_1^2 x^2 dx)(\int_1^2 x^2 dx)$$. Proceed to integrate and get $$I^2 = (\frac{1}{3}(8-1))(\frac{1}{3}(8-1)) = \frac{49}{9}$$. Now take the square root of each side and get $$I = \sqrt{\frac{49}{9}} = \frac{7}{3}$$. Now calculate I and square your answer. You'll get $$\frac{49}{9}$$.

* I choose the positive root of the when I took the square root since we know that $$\int_1^2 x^2 dx = \frac{7}{3}$$.

 

[dextercioby]

Of course, one would first to show that the <squared> integral is not divergent. Only then you can <square> it. And one more thing, there's a theorem of Fubini whose simplest application is the <squaring> mentioned above.

 

[Samuelb88]

If $$f(x)$$ is continuous of [a,b], then it's integral from a to b exists, right?

 

[cragar]

They do it a little bit differently with the Gaussian integral. When they square it they turn one into a y and then combine them, then do it in polar. I don't know it just seems weird.
http://en.wikipedia.org/wiki/Gaussian_integral
It shows it in the second part.

 

[Samuelb88]

Yes, but $$\int_0^{\infty} e^{-x^2} dx$$ is the same thing as $$\int_0^{\infty} e^{-y^2} dy$$. Hence: $$(\int_0^{\infty} e^{-x^2} dx)(\int_0^{\infty} e^{-x^2} dx) = (\int_0^{\infty} e^{-x^2} dx)(\int_0^{\infty} e^{-y^2} dy)$$.

 

[cragar]

Yes, but $$\int_0^{\infty} e^{-x^2} dx$$ is the same thing as $$\int_0^{\infty} e^{-y^2} dy$$. Hence: $$(\int_0^{\infty} e^{-x^2} dx)(\int_0^{\infty} e^{-x^2} dx) = (\int_0^{\infty} e^{-x^2} dx)(\int_0^{\infty} e^{-y^2} dy)$$.

then they combine the x and y pieces to have

$$\int\int e^{-x^2-y^2}dxdy$$
I don't know maybe its ok.

 

[micromass]

then they combine the x and y pieces to have

$$\int\int e^{-x^2-y^2}dxdy$$
I don't know maybe its ok.

It's just doing

$$a\int{e^{x^2}dx}=\int{ae^{x^2}dx}$$

but with

$$a=\int{e^{y^2}dy}$$

 

[Hurkyl]

then they combine the x and y pieces to have

$$\int\int e^{-x^2-y^2}dxdy$$
I don't know maybe its ok.

After mulling over the operations, you should be able to convince yourself that getting to this point is rock solid.

The main thing to worry about for this proof is the change of variable from Cartesian coordinates to polar coordinates. Something needs to be said here. (and as an aside, I once read an article suggesting there was a very narrow class of integrands where it is actually legitimate to use this overall trick)

 

[Char. Limit]

In order to square an integral, integrate, and then take the square root, don't you first have to show that the integral (or perhaps the integrand) is positive over the region of integration?

 

[cragar]

when we square it why can we change one of the x's to a y? Now were integrating over a different axis.

 

[HallsofIvy]

We are not "squaring" it. As has been said, If
$$I= \int_0^\infty e^{-x^2}dx$$
then it is also true that
$$I= \int_0^\infty e^{-y^2}dy$$

Now multiply the left and right sides of those two equations:
$$I(I)= I^2= \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy\right)$$

Then Fubini's theorem says that the product of integrals can be written as a double integral:
$$I^2= \int_{x=0}^\infty \int_{y=0}^\infty e^{-(x^2+ y^2)}dydx$$.

 

[cragar]

your good