text: Q:
  
  A doubt about a trivial Dirac Delta Integral
  
  The following is, in my opinion, a quite simple integral. Yet I can figure out why the result is thy professor said to be.
  The integral is:
  $$\int_0^{+\infty} \delta(\sin(\pi x))2^{-x}\ dx$$
  By Dirac Delta properties:
  $$\delta(\sin(\pi x)) = \sum_{k = 0}^{+\infty}\frac{\delta(x-k)}{|\pi \cos(\pi x)|_{x = k}}$$
  For $k\in\mathbb{N}$ since the integral is over $\mathbb{R}^+$.
  Hence the cosine is $1$ and we have
  $$\sum_{k = 0}^{+\infty} \frac{1}{\pi}\int_0^{+\infty}\delta(x-k)2^{-x}\ dx$$
  $$\sum_{k = 0}^{+\infty} \frac{2^{-k}}{\pi} = \frac{2}{\pi}$$
  Yet my professor says it has to be
  $$\frac{2}{3\pi}$$
  How is that possible?
  EDITS
  Ok, understood my stupid mistakes about
  $$\cos(\pi k) = (-1)^k$$
  The question now is: there is the absolute value, hence shouldn't 
  $$|(-1)^k|$$
  Be always positive??
  Anni the additional factor, the series would become
  $$\sum_{k = 0}^{+\infty} (-1)^k 2^{-k} = \frac{2}{3}$$
  
  A:
  
  Your series expansion for $\delta(\sin\pi x)$ is correct. I think your problem is because you are integrating a $\delta(x)$ from $x=0$. We write $H(x):=\int_{-\infty}^x\delta(\xi)\mathrm d\xi$ for the Heaviside step function. Now we know that for $\epsilon>0$, we have $\int_{-\epsilon}^\epsilon\delta(x)f(x)\mathrm d x = f(0)$ but one part of your integral (for the term where $k=0$) is $\int_0^\infty\frac{\delta(x)}\pi2^{-x}\mathrm d x$ which you claim is $\frac1\pi$ and I claim is $\frac{1-H(0)}\pi.$ And thus I think your professor is using the convention that $H(0)=\frac12.$ Then your answer would be too big by $\frac1{2\pi},$ which it is.
  This correction gives the answer as:
  $$\frac1\pi\left(1-H(0)+\sum_{k=1}^\infty2^{-k}\right)$$
  Thus you get $\frac1\pi\left(1+1\right)$ and I get $\frac1\pi\left(\frac12+1\right)=\frac3{2\pi}$, the same as your professor.
  Edit: You’ve changed your question to say that your professor thinks the solution is $\frac2{3\pi}.$ This is not the solution.
  Further details:
  What is $\int_0^\infty\delta(x)\mathrm d x$? Well by the definition of $H$ above we have:
  $$\int_0^\infty\delta(x)\mathrm d x =\left[H(x)\right]_0^\infty = H(\infty) - H(0) = 1 - H(0).$$ we leave $H(0)$ unevaluated because it is usually a convention. Some people choose $H(0)=1$, others $H(0)=0$ and others still choose $H(0)=\frac12$. I have heard rumours of some particularly applied physics calculations which depend on the value of $H(0)$ and required $H(0)=\frac12$ to get the right answer. Another vote in its favour is that it is then most “almost continuous” option in the following sense: It gives us that $H(x)=\frac12\left(\lim_{\xi\to x^-}H(\xi) + \lim_{\xi\to x^+}H(\xi)\right),$ the same property you get after a forward then inverse Fourier transform of a function.
  
  