In Question (2.20) of Griffiths' Quantum Mechanics book, they have given this Solution.

In the Solution of question 2.20(b), they omitted $e^{(ik-a) \infty}$ (or may have considered $e^{(ik-a) \infty}=0$) in this calculation

calculation of 2.20(b).

How can it be correct at all?

  • 8
    $\begingroup$ That’s not $e^{\infty}$, it’s $e^{-\infty}$ which is zero. $\endgroup$
    – Zack
    May 1, 2022 at 17:44

1 Answer 1


You can rewrite the indefinite integral as \begin{equation} e^{-a x} f(x) \end{equation} where $f(x)$ does not grow exponentially with $x$. (In fact $f(x) \sim e^{\pm i k x}$ is an oscillating function).

In the limit $x\rightarrow \infty$, we have that $e^{-a x} f(x) \rightarrow 0$, since $e^{-ax} \rightarrow 0$ and $f(x)$ doesn't grow fast enough to cancel this behavior. This assumes ${\rm Re}( a )> 0$.

  • 1
    $\begingroup$ Another way of saying this is that $\operatorname{Re}(a)>0$, is that makes the integral absolutely convergent, so we can treat it as just a Riemann integral. Then, the bracket at the end is : $$\left[\frac{e^{(ik-a)x}}{ik-a}\right]_0^\infty= \left(\lim_{x\to +\infty} \frac{e^{(ik-a)x}}{ik-a}\right)- \frac{1}{ik-a} = -\frac{1}{ik-a}$$ because $\lim_{x\to +\infty} e^{(ik-a)x} = 0$ $\endgroup$ May 1, 2022 at 19:01

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