# Integration help with normalizing wave function

1. Jan 12, 2014

### leroyjenkens

1. The problem statement, all variables and given/known data
ψ(x,t) = Ae-λxe-iwt

Normalize this and solve for A

2. Relevant equations
$$\int_{-∞}^{∞}|ψ|2dx = 1 3. The attempt at a solution I got to A2$\int_{-∞}^{∞}e^{-2λx}dx$ The solution manual multiplies it by 2 and only goes from 0 to ∞ instead of from -∞ to ∞. I'm trying to do it from -∞ to ∞ and I should get the same answer as the manual, but I'm not. They get a finite non-zero answer, but integrating what I got to, I have $-\frac{A^{2}}{2λ}e^{-2λx}$ and now, plugging in the limits of integration in for x, it will give me infinity as the answer. Anyone see where I made a mistake? Thanks. 2. Jan 12, 2014 ### jbunniii The integral as written above is infinite. It is not the same as integrating from $0$ to $\infty$ and multiplying by $2$. Just plot the integrand to see this. If the integrand were $e^{-2\lambda |x|}$ (with absolute values) then it would be true. 3. Jan 12, 2014 ### leroyjenkens It is supposed to be absolute value. I keep forgetting about that. But now I'm getting 0 as the answer, since the ∞ in the exponent will make it look like this when I evaluate the integral over the limits: [tex]-\frac{A^{2}}{2λ}[e^{-2λ∞}-e^{-2λ∞}]$$

Which is just

$$-\frac{A^{2}}{2λ}[0 - 0]$$

Which equals 0.

I know I'm doing something wrong, though.

Thanks for the response.

4. Jan 12, 2014

### jbunniii

What integral did you perform? What were the upper and lower limits?

5. Jan 12, 2014

### leroyjenkens

Upper limit was ∞ and lower was -∞. I plug them in for x, and since it's the absolute value of x, they're both just ∞, which means it turns to 0.

6. Jan 13, 2014

### jbunniii

So you are trying to perform this integral:
$$\int_{-\infty}^{\infty} e^{-2 \lambda |x|} dx$$
Because of the absolute value, you can't evaluate the integral without breaking it into parts. (If you believe otherwise, then you need to produce an antiderivative for $e^{-2\lambda|x|}$.) Try writing it this way:
$$\int_{-\infty}^{\infty} e^{-2 \lambda |x|} dx = \int_{-\infty}^{0} e^{-2 \lambda |x|} dx + \int_{0}^{\infty} e^{-2 \lambda |x|} dx$$
Next step: simplify the integrand in each case by replacing $|x|$ with something simpler.