We don't have to do any way of the calculation, but this is clear enough.
So we're integrating first in x and then the inner integral will be in t because we want
to exploit here the integration by parts formula in time, that's what we would like to do.
I will need to rewrite things a little.
So, what is the definition of u? As long as t is larger than x over b, so it depends
also on where is x. The point is that if x is negative, t is always larger than x over
b. So as long as x moves from minus infinity to 0, t is always larger and so we have only
one definition. We have tb minus x against phi t in dt dx. Plus, second case, if x is
positive, then we have that either t is less than x over b, so t from 0 to x over b and
in this case we have x minus tb phi t in dt plus change sign. So from x over b to plus
infinity tb minus x against phi t in dt, all of these in dx. And this is only for this
part. We are analyzing this term. So let's do the integration by parts in time. Here
we get minus infinity 0. Well, again I told you phi plus infinity is 0, so there's nothing
to say. Phi in 0 is a number, but here it gives us minus x, so minus minus plus x phi
x0 and then minus the integral from 0 to plus infinity of, well, the derivative of this
give us b phi in dt and all of this in dx plus, also here, the integral from 0 to plus infinity.
When I integrate my part of this term, I get, again, when t is equal to x over b, this vanishes,
when t is equal to 0 instead I get x, the minus phi x0 minus the integral from 0 to x over b of the
derivative, so the derivative in t gives minus b, so minus minus plus b phi in dt. And this
term, integration by parts, so the evaluation at plus infinity, of course, is 0 because phi is
complex apart. When t is equal to x over b, this term vanishes. So we get only the second integral
from x over b to plus infinity minus and the derivative of this gives b phi in dt. All of
these in dx. Let's put things together. What is written here? Here is written x phi x0 when x is
negative and here minus x phi x0 when x is positive, so this is minus the absolute value of x,
which is minus g1, so we have minus the integral over r of absolute value of x phi x0 in dx.
And that's beautiful that we found because this is the absolute value of x. This term, which is
coming from here, will be simplified with the term coming from there. We are on the right track.
The other terms should hopefully vanish when we expand this other part of the integral.
The other pieces is that we can also put together. So here we have minus the integral from minus
infinity to 0 in x times the integral from 0 to plus infinity in time. And here we have the other
part of the integral, so in x positive. And what do we have? But I guess I should leave them like this.
Minus the integral from minus infinity to 0 times the integral from 0 to plus infinity of b
phi dt dx plus the integral from 0 to plus infinity of the integral from 0 to x over b
of b phi dt dx minus the integral from 0 to plus infinity of the integral from x
over b to plus infinity of b phi dt dx so as I said now the whole point will be to
show that the integral of u against b phi x is equal to minus these
three terms and as you may have imagined this will come by symmetry because I
mean we again will use the same trick but let's see of course