How Does the Integral of exp(-x^2*k+i*m*x) Behave as m Increases?

[ersanjaynp]

$$\int ^{5}_{0} \left cos(m*x)e^{- \frac{x^{2}}{a^{2}}} \right dx$$

 

[arildno]

Well, you can first the integrand as:
$$\frac{1}{2}(e^{mix-\frac{x^{2}}{a^{2}}}+e^{-imx-\frac{x^{2}}{a^{2}}}), i=\sqrt{-1}$$
Complete the squares in the exponents to proceed a bit further.

 

[ersanjaynp]

Well, you can first the integrand as:
$$\frac{1}{2}(e^{mix-\frac{x^{2}}{a^{2}}}+e^{-imx-\frac{x^{2}}{a^{2}}}), i=\sqrt{-1}$$
Complete the squares in the exponents to proceed a bit further.

how to integrate the first part only i.e. exp(-x^2*k+i*m*x) for some defined limits, say 0 to b. Mathcad goves solution in the form of complex error function that is difficult to define as erfi (m) when m is large can't be defined (infinite)..but the function has zero values at higher m...

 

[elibj123]

how to integrate the first part only i.e. exp(-x^2*k+i*m*x) for some defined limits, say 0 to b. Mathcad goves solution in the form of complex error function that is difficult to define as erfi (m) when m is large can't be defined (infinite)..but the function has zero values at higher m...

The primitive function of exp{-x^2} is not elementary, therefore cannot be expressed in simpler terms than the error function which is numerically approximated.

As for the values of the integral as m tends to large numbers, this is an instance of the Riemann-lebesgue lemma, which generally says that for any smooth function f(t) the integral $$\int_{interval}cos(wt)f(t)dt$$ will tend to zero as w tends to infinity