# Thread: How did he jump from this to that

1. ## How did he jump from this to that

How do i get from this $\displaystyle 1/2\pi*\int_{-\infty}^\infty\ dt \int_{-3\pi}^{3\pi}\ sin(u)sin(tx)sin(tu) du$

to this:

$\displaystyle 2/\pi*\int_{0}^\infty\ dt\int_{0}^{3\pi}\sin(u)sin(tx)sin(tu) du$

How and why does he change the intervals? I can't understand what he did.
Could someone please explain?

It's not multiplied, it's a double integral.

2. Here's a thought, if a function is symmetrical around 0 then a definite integral with symmetrical terminals can be changed if the intgeral is doubled.

I.e $\displaystyle \displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x~dx =2\int_{0}^{\frac{\pi}{2}} \cos x~dx$

Does this help?

3. So you're baiscally saying that since the area is the same on both sides of the axis, the notation is equal to us simply doubling the integral, which does make sense.

In order for a function to be simmetric towards 0, it means it has to be even. Am i wrong on that? So since we have the product of two sines, it means it will result in an even function.