# Trouble solving an integral

So I have been trying to solve this equation,

$\psi(E) = \int_0^\infty \psi_0 exp(\frac{iE_0 t}{h} - \frac{iEt}{h} - \frac{t}{\Delta t})dt$

$\psi(E) = \frac{-i \psi_0 h \Delta t}{(E_0 - E)\Delta t + ih}$

I began by using substitution to change the integral.

$u = \frac{iE_0 t}{h} - \frac{iEt}{h} - \frac{t}{\Delta t}$

$\psi(E) = \int \psi_0 \frac{1}{\frac{iE_0 }{h} - \frac{iE}{h} - \frac{1}{\Delta t}} exp(u)du = \psi_0 \frac{h\Delta t}{iE_0 \Delta t- iE \Delta t- h} \int exp(u)dt = \psi_0 \frac{h\Delta t}{iE_0 \Delta t- iE \Delta t- h} exp(u)$

Substituting t back in

$\psi_0 \frac{h\Delta t}{iE_0 \Delta t- iE \Delta t- h} exp(\frac{iE_0 t}{h} - \frac{iEt}{h} - \frac{t}{\Delta t})$

where t is taken from 0 to infinity.

Now if I take this answer minus the exponential, I can rearrange it to get the given answer

$\psi_0 \frac{h\Delta t}{iE_0 \Delta t- iE \Delta t- h} = \frac{\psi_0 h\Delta t}{i \Delta t (E_0- E) - h} = \frac{i\psi_0 h\Delta t}{- \Delta t (E_0- E) - ih} = \frac{-i\psi_0 h\Delta t}{ \Delta t (E_0- E) + ih}$

Which would mean that the exponential that this is multiplied by (taken from t = 0 to t = $\infty$) would have to equal 1. I'm having trouble with the exponential, but I can't think of any way that it could be equal to 1.

Where did I go wrong? I feel like I've made a stupid mistake somewhere, but I can't seem to find it.

• While it's easy to point out where you went wrong, the bigger issue is the way you go about doing the computations, which makes it hard for you to see where you went wrong. It's better to define a single symbol for the coefficient of t in the exponential, write it as exp(-a t) and then show that the integral from 0 to infinity in case the real part of a is positive is given by 1/a. Then you substitute for a the desired expression. – Count Iblis Apr 26 '15 at 17:19

The exponential term can be slightly rearranged as

$$e^{-(1/\Delta t-i(E_0-E)/h)t}=e^{-(1/\Delta t)t}e^{i(E_0-E)/ht}$$

Taking the magnitude of the right-hand side and exploiting the fact that for real-valued $x$, $|e^{ix}|=|\cos x + i \sin x|=\sqrt{\cos^2x+\sin^2x}=1$, we find

$$0\le |e^{-(1/\Delta t)t}e^{i(E_0-E)/ht}|\le e^{-(1/\Delta t)t}$$

and the right-hand side approaches $0$ when $t \to \infty$.

Now, the integral becomes easy to evaluate as

\begin{align}\int_0^{\infty} \psi_0 e^{-(1/\Delta t-i(E_0-E)/h)t}dt&=-\frac{\psi_0}{1/\Delta t-i(E_0-E)/h}\left(e^{-(1/\Delta t)t}e^{i(E_0-E)/ht}\right)|_0^{\infty}\\\\ &=\frac{\psi_0}{1/\Delta t-i(E_0-E)/h}\\\\ &\frac{i\psi_0h\Delta t}{(E_0-E)\Delta t+ih} \end{align}

as expected!!

• Two questions. First, why is the absolute value of the original exponential less than (or equal to) the exponential on the right? Second, wouldn't the original exponential as t goes to 0 then need to be -1 (exp. as t is taken to infinity minus exp. as t is taken to 0, 0 - (-1) = +1), which isn't possible? – Student Apr 26 '15 at 18:12
• Sure. For the first question, I added an explanation in the posted answer. The reason is that for real values of $x$, $|e^{ix}|=1$. Both exponentials are equal to $1$ when $t=0$. Recall that the anti-derivative is negative and so you flip the order of upper and lower limits. – Mark Viola Apr 26 '15 at 18:26
• There's a negative sign in front of the given answer. – Student Apr 26 '15 at 19:33
• The given answer is incorrect. Your is correct. – Mark Viola Apr 26 '15 at 20:26