# 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

#### Prove It

MHF Helper

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?

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

Im still sure the op is wrong any help?

#### drumist

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.

Notice that $$\displaystyle (a-b)^2 = (b-a)^2$$ for any $$\displaystyle a,b$$.
So $$\displaystyle \frac{\lambda^2}{(\lambda - i t)^2} = \frac{\lambda^2}{(it-\lambda)^2}$$