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I am trying to evaluate the integral $$\int_{-\infty}^{\infty} f(x) \delta(ax-b) \, dx$$ for $a\neq0$. From what I was taught, I would expect the answer to be $f\bigl(\frac{b}{a}\bigr)$ since $a\cdot\frac{b}{a}-b=0$.

However, I can also do a change of variables $u=ax$. This makes the integral $$\frac{1}{a}\int_{-\infty}^{\infty} f \Bigl(\frac{u}{a}\Bigr) \delta(u-b) \, du.$$ I would expect the value of this expression to be $\frac{1}{a} f\bigl(\frac{b}{a}\bigr)$. How am I getting two different answers? Is one of my steps invalid? Wolfram|Alpha seems to agree with the second result. If this is correct, why do I have to make the substitution to find the correct answer?

Thank you

MathIsFun7225
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  • There's no such rule which states that if we consider: $$I=\int_{-\infty}^{\infty} f(x)\delta(\cdot)~dx$$ and if the function inside the Dirac delta function has a root at a point $x_1$ then the result is $I=f(x_1)$. – projectilemotion Jan 24 '19 at 22:57

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Your original solution is indeed false. Because $\delta$ isn't a function, you have to be a little bit more careful on what you can and cannot do. If you scale the "argument" you will indeed get this factor. Depending on how far into the theory you want to go you can show (or accept as a definition) that for $a \neq 0$

$$\int_{\mathbb R} f(x)\delta(ax)dx = \frac{1}{|a|} \int_{\mathbb R} f(x/a)\delta(x)dx.$$

(Check out the properties of the dirac delta distribution.)

flawr
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