# Math Help - Can you check this integration for me please

1. ## Can you check this integration for me please

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

ip means integration by parts

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

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

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

Now using integration by parts with

$u = x$ so that $du = 1$

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

we find

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

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

Can you finish?

3. Originally Posted by Prove It
If you are trying to evaluate $\lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx}$

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

Now using integration by parts with

$u = x$ so that $du = 1$

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

we find

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

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

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

4. Im still sure the op is wrong any help?

5. Notice that $(a-b)^2 = (b-a)^2$ for any $a,b$.

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

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

6. Originally Posted by drumist
Notice that $(a-b)^2 = (b-a)^2$ for any $a,b$.

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

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