Can anyone solve the problem given in the attachment? here a, b and 2pi are constants. Please show the steps while u solve it.
Bye.
What did you get so far and is this what it looks like?
$\displaystyle \frac{1}{a2\pi}\int_{-\infty}^{\infty} ye^{(\frac{-y^2}{b})}~dy$
$\displaystyle \frac{-be^{(\frac{-y^2}{b})}}{a4\pi}\bigg|^{\infty}_{-\infty}$
Well I want to test the previous post. If it's true then i'm going to have a lot less work to do!
$\displaystyle \frac {1}{a(2 \pi)}\int_{-\infty}^{+\infty}ye^{ \frac {-y^2}{b}}dy$
$\displaystyle u=\frac {-y^2}{b}$
$\displaystyle \frac {du}{dy}=\frac{-2y}{b}$
$\displaystyle dy= \frac {b du}{-2y}$
The limits remain the same. I do not think it is necessary to write that part of the working out.
$\displaystyle \frac{1}{a(2 \pi)(-2)}\int_{- \infty}^{\infty} \frac {y}{y}udu$
$\displaystyle \frac{1}{-4a \pi}\int_{- \infty}^{\infty} udu$
$\displaystyle \frac {1}{-4a \pi} [\frac{u^2}{2}]^{\infty}_{-\infty} $
$\displaystyle \frac {1}{-8a \pi}[u^2]^{\infty}_{-\infty}$
$\displaystyle \frac {1}{-8a \pi}((\infty)^{2}-(-\infty)^{2})$
$\displaystyle \frac {1}{-8a \pi}(0)=0$
WOW! That is the find of the century for me!!!!
(oh, and that's how I would integrate it.)
showcase, you have two mistakes in your arguments -- namely, with "the limits stay the same", and after the change of variables you lost exponential for some reason.
basically 11rdc11 wrote out the solution -- the primitive is $\displaystyle \frac{-be^{(\frac{-y^2}{b})}}{a4\pi}$ and then the integral is $\displaystyle \lim_{y\to\infty}\frac{-be^{(\frac{-y^2}{b})}}{a4\pi}-\lim_{y\to -\infty}\frac{-be^{(\frac{-y^2}{b})}}{a4\pi}=0$.
the short way of this, as I said, is to note that the function is odd, so the integral is zero immediately, though I am sloppy here, because for this argument to work, one needs to check that the integral is finite at first