Can you check this integration for me please

Oct 2008
393
10


I think the denominator should be the other way around: it-lambda

ip means integration by parts
 

Prove It

MHF Helper
Aug 2008
12,883
4,999


I think the denominator should be the other way around: it-lambda

ip means integration by parts
If you are trying to evaluate \(\displaystyle \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx}\)


\(\displaystyle \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx} = \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t - \lambda)e^{x(i\,t - \lambda)}\,dx}\)

Now using integration by parts with

\(\displaystyle u = x\) so that \(\displaystyle du = 1\)

and \(\displaystyle dv = (i\,t - \lambda)e^{x(i\,t - \lambda)}\) so that \(\displaystyle v = e^{x(i\,t - \lambda)}\)

we find

\(\displaystyle \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t - \lambda)e^{x(i\,t - \lambda)}\,dx} = \left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \int{e^{x(i\,t - \lambda)}\,dx}\right]_0^{\varepsilon}\)

\(\displaystyle =\left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \frac{e^{x(i\,t - \lambda)}}{i\,t - \lambda}\right]_0^{\varepsilon}\)

Can you finish?
 
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Oct 2008
393
10
If you are trying to evaluate \(\displaystyle \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx}\)


\(\displaystyle \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx} = \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t - \lambda)e^{x(i\,t - \lambda)}\,dx}\)

Now using integration by parts with

\(\displaystyle u = x\) so that \(\displaystyle du = 1\)

and \(\displaystyle dv = (i\,t - \lambda)e^{x(i\,t - \lambda)}\) so that \(\displaystyle v = e^{x(i\,t - \lambda)}\)

we find

\(\displaystyle \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t - \lambda)e^{x(i\,t - \lambda)}\,dx} = \left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \int{e^{x(i\,t - \lambda)}\,dx}\right]_0^{\varepsilon}\)

\(\displaystyle =\left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \frac{e^{x(i\,t - \lambda)}}{i\,t - \lambda}\right]_0^{\varepsilon}\)

Can you finish?
Thanks i don't know how to use latex so paint will have to do lol

 
Jan 2010
354
173
Notice that \(\displaystyle (a-b)^2 = (b-a)^2\) for any \(\displaystyle a,b\).

So \(\displaystyle \frac{\lambda^2}{(\lambda - i t)^2} = \frac{\lambda^2}{(it-\lambda)^2}\)

If you need convincing, just try to square both of the denominators and see that you get the same result.
 
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Oct 2008
393
10
Notice that \(\displaystyle (a-b)^2 = (b-a)^2\) for any \(\displaystyle a,b\).

So \(\displaystyle \frac{\lambda^2}{(\lambda - i t)^2} = \frac{\lambda^2}{(it-\lambda)^2}\)

If you need convincing, just try to square both of the denominators and see that you get the same result.
Thank you (Clapping)