Straightforward integral?

1. Jan 25, 2007

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

2. Jan 26, 2007

HallsofIvy

$$\int e^{a k_r}dr= \frac{1}{a}e^{ak_r}$$
but you have kr2. That square in the exponential means that the anti-derivative cannot be written as an elementary function.
Are you familiar with
$$\int_{-\infty}^\infty e^{-x^2}dx= \sqrt{\pi}$$?