Your description of Delta function is not correct. Delta(t-c)=infinity at x=c, and =0 at x which is not equal to c. Then the integral of Delta(t-c) is Zero by Lebesgue inetegral theory. You had better study the distribution theory by Laurent Schwartz and Lebesgue integral theory.
I do not think that the argument to show that the integral of the Dirac delta function is 1 using limits of a function is a valid proof. If you replace the 2 in the function that has a magnitude of 1/2e you will still approach the same limit. meaning that a function G(t) defined in the range -e to e with a magnitude of 1/3e gives an area of 2/3. Now the limit as e goes to zero 1/3e goes to infinity. Therefore using the same argument, the integral over the infinite range of the Dirac delta function is 2/3. Generalizing, If I had any function defined in the range -e to e with magnitude 1/ne where n is any integer, then the integral over the infinite range could be any number (2/n). Therefore if the argument given is valid, the integral does not exist.
Find other Differential Equations videos in my playlist ua-cam.com/play/PLkZjai-2JcxlvaV9EUgtHj1KV7THMPw1w.html
Thank you so much, I was solving David J.griffths book of electrodynamics this is helpful for me.
Thanks for your great explanation about Dirac Delta Function!
Nice review for Dirac delta function,thank you very much
Your description of Delta function is not correct.
Delta(t-c)=infinity at x=c, and =0 at x which is not equal to c.
Then the integral of Delta(t-c) is Zero by Lebesgue inetegral theory.
You had better study the distribution theory by Laurent Schwartz and Lebesgue integral theory.
Your voice is quite pleasant to hear.....♥️♥️♥️
I do not think that the argument to show that the integral of the Dirac delta function is 1 using limits of a function is a valid proof.
If you replace the 2 in the function that has a magnitude of 1/2e you will still approach the same limit. meaning that a function G(t) defined in the range -e to e with a magnitude of 1/3e gives an area of 2/3.
Now the limit as e goes to zero 1/3e goes to infinity. Therefore using the same argument, the integral over the infinite range of the Dirac delta function is 2/3.
Generalizing, If I had any function defined in the range -e to e with magnitude 1/ne where n is any integer, then the integral over the infinite range could be any number (2/n).
Therefore if the argument given is valid, the integral does not exist.
You can't argue about a definition.
Thank you👍
really extremely helpful! thank you sir you made my day lol!
Great video. Thanks
Is dirac function and its properties rigorous according to mathematics??
There is mathematical rigor. It is called a distribution. Not many physicists care about that.
@@ProfJeffreyChasnov thanks for the fast reply!!
Awesome!!!
Please solve examples
Look at Lecture 35 in the playlist.