Dirac delta and exponential

[eljose]

Let be the exponential:

$$e^{inx}=cos(nx)+isin(nx)$$ $$n\rightarrow \infty$$

Using the definition (approximate ) for the delta function when n-->oo

$$\delta (x) \sim \frac{sin(nx)}{\pi x}$$ then differentiating..

$$\delta ' (x) \sim \frac{ncos(nx)}{\pi x}- \frac{\delta (x)}{\pi x}$$

are this approximations true for big "n" ??.. i would like to know this to compute integrals (for big n ) of the form:

$$\int_{2}^{\infty} dx f(x,n)e^{inx}$$

thanks... :grumpy: :tongue2:

 

[WigneRacah]

You can say:

$$\frac{\sin(nx)}{\pi x} \rightarrow \delta (x)$$

for $$n \rightarrow \infty$$ in the sense of distributions. But, by derivating, you have:

$$\frac{n x \cos(n x) - \sin(n x)}{\pi x^2} \rightarrow \delta'(x)$$.

$$\frac{ \cos (n x)}{x^2}$$ and $$\frac{1}{x} \delta(x)$$ are not distributions! They are meaningless as distributions.

 

[tacman]

The approximations for the Dirac delta function and exponential are indeed true for large values of n. As n approaches infinity, the exponential term becomes more dominant and the Dirac delta function becomes more concentrated at the origin. This means that the approximation for the delta function becomes more accurate.

Using this approximation, we can compute integrals involving the exponential term for large values of n. The integral provided in the question, \int_{2}^{\infty} dx f(x,n)e^{inx}, can be evaluated using the approximation for the delta function. This can be done by rewriting the integral as \int_{2}^{\infty} dx f(x,n)\frac{sin(nx)}{\pi x} and then taking the limit as n approaches infinity.

In general, the Dirac delta function and exponential are closely related and their approximations can be used to simplify calculations in various mathematical and scientific fields. However, it is important to note that these approximations are only valid for large values of n and may not hold for smaller values. It is always best to check the validity of the approximation before using it in any calculations.