If the given integral is $$\int_{-\infty}^{+\infty} dx \delta (x-x^{'})f(x)$$ The answer is $f(x')$. However if we make a transformation$$x\rightarrow\alpha x=y$$and $$x^{'}\rightarrow\alpha x^{'}=y^{'}$$ substituting this into the above equation I get$$\alpha dx= dy$$

so the integral becomes, $$\frac{1}{\alpha}\int dy \delta (\frac{y}{\alpha}-\frac{y^{'}}{\alpha})f(\frac{y}{\alpha} )$$ Hence the answer is $$\frac{1}{\alpha}f(\frac{y^{'}}{\alpha})$$

If we go back to the old variables $x$ by making an inverse transformation $$y^{'}=\alpha x^{'}$$ I am getting $$\frac{1}{\alpha}f(x^{'})$$ However the solution is not just $$f(x^{'})$$ Where am I going wrong in the following transformation ?


Your error lies in assuming that $$\int \delta(x - b ) f(x) dx = f(b) \implies \int \delta(ax - b ) f(x) dx = f(b/a)$$ i.e. that the delta integral merely "picks the value of the function where the arg of the delta is zero"

Actually scaling the delta argument scales the result. In particular, for any $a>0$

$$ \int \delta(ax) dx = \frac{1}{a} $$

and also $$ \int\delta(ax - b ) f(x) dx = \frac{1}{a} f(b/a) $$

Hence $$ \frac{1}{\alpha}\int dy \delta (\frac{y}{\alpha}-\frac{y^{'}}{\alpha})f(\frac{y}{\alpha} ) = f(\frac{y^{'}}{\alpha})$$

  • $\begingroup$ Thank you so much for your explanation. $\endgroup$ – user135580 Jun 3 '17 at 18:32

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