What is the answer of this indefinite integral containing a delta and an exponetial functions?

Question

In during an integral solving by employing Fourier transform, below integral appeared $$ \int_{-\infty}^{\infty} \frac{\delta(k-\beta)}{k}exp(-\alpha^2(k+b)^2)dk$$. What is its answer? I gauss it must be equal $$ =\frac{1}{\alpha}e^{-\alpha^2(\beta+b)^2)} $$ when $\beta=k$. But based on my professor opinion, its answer is $$ =\frac{1}{\alpha\beta}e^{-\alpha^2(\beta+b)^2)} $$. Because when $\beta=k$ forced to appear $\beta$ in the denominator. I need help, please.

Answer 1

Well, you can use the following relation:

$$\mathcal{I}_\text{n}:=\int_\mathbb{R}\delta\left(x-\text{n}\right)\text{y}\left(x\right)\space\text{d}x=\text{y}\left(\text{n}\right)\tag1$$

For $\text{n}\in\mathbb{R}$.

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So, for your case we get:

$$\int_\mathbb{R}\delta\left(x-\beta\right)\cdot\frac{\exp\left(-\alpha^2\cdot\left(x+\text{b}\right)^2\right)}{x}\space\text{d}x=\frac{\exp\left(-\alpha^2\cdot\left(\beta+\text{b}\right)^2\right)}{\beta}\tag2$$

For $\beta\in\mathbb{R}$.