# How can we ignore the diverging term $e^\infty$ in the integral?

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

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How can it be correct at all?

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

You can rewrite the indefinite integral as $$$$e^{-a x} f(x)$$$$ where $$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$$.
• Another way of saying this is that $\operatorname{Re}(a)>0$, is that makes the integral absolutely convergent, so we can treat it as just a Riemann integral. Then, the bracket at the end is : $$\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 $\lim_{x\to +\infty} e^{(ik-a)x} = 0$ May 1, 2022 at 19:01