Straightforward integral?

[Bazman]

Hi,

i've seen a textbook where the integration inthe following expression is performed:


C(S,V,t) = 1-(1/2pi)*exp[-lambda(k0)*t](u(k0,V)/k0^2-i*k0)
+ infin
*S exp[-.5*kr^2*lambda''(k0)*t]dkr
- infin

where kr is k subscript r and k0 is ksubscript 0
i = complex number
lamda'' =2nd derivative of lambda

giving


=1-(u(k0,V)/(k0^2-i*k0)*(1/(2pi*lamda''(k0)*t)^.5)*exp[-lambda(k0)*t]


to me the integral

+ infin
*S exp[-.5kr^2*lambda''(k0)t]dkr
- infin

should give exp[-.5kr^2lambda''(k0)*t]/(-kr*lambda''(k0)*t)

evaluated at + and - infin and in both cases should go to zero but from the solution above this is clearly incorrect

please let me know where I am going wrong

 

[HallsofIvy]

$$\int e^{a k_r}dr= \frac{1}{a}e^{ak_r}$$
but you have k_r². That square in the exponential means that the anti-derivative cannot be written as an elementary function.
Are you familiar with
$$\int_{-\infty}^\infty e^{-x^2}dx= \sqrt{\pi}$$?

 

[tacman]

Hi there,

Thank you for sharing your thoughts on this integral. It seems like there may be a few misunderstandings about the integration process in this expression. Let's break it down together:

First, the expression C(S,V,t) is defined as a function of three variables: S, V, and t. This means that when we integrate it, we need to specify which variable we are integrating with respect to. In this case, it seems like the integration is being done with respect to ksubscript r, which is represented as "kr" in the integral.

Next, the integral itself is a definite integral, meaning it has limits of integration. In this case, the limits are from -infin to infin. This means that we are integrating over all possible values of kr, from negative infinity to positive infinity.

Now, let's look at the integrand (the expression being integrated). It contains two terms: the first term is a constant, and the second term is a function of kr. The first term can be pulled out of the integral since it does not depend on kr. The second term, however, cannot be pulled out of the integral and must be integrated separately.

To integrate the second term, we can use the fact that it is in the form of a Gaussian function (exp[-.5kr^2*lambda''(k0)t]). This type of function can be integrated using the Gaussian integral formula, which is:

int(exp[-.5x^2])dx = sqrt(2*pi)

In our case, we have exp[-.5kr^2*lambda''(k0)t], so we can replace x with kr and evaluate the integral from -infin to infin. This gives us:

sqrt(2*pi/lambda''(k0)t)

Now, we can combine this with the first term that we pulled out of the integral, which gives us:

1-(1/2pi)*exp[-lambda(k0)*t](u(k0,V)/k0^2-i*k0)*(sqrt(2*pi/lambda''(k0)t))

This is equivalent to the solution given in the textbook. It is important to note that the limits of integration are not evaluated at infinity, but rather the integral is evaluated from -infin to infin and then multiplied by 2 to account for the entire range of integration.

I hope this explanation helps clear up any confusion. Please let me know if