2

A generalization of the Gaussian integral $$ \int_0^{\infty} e^{-\pi at^2} \, dt = \frac{1}{2\sqrt{a}}, \hspace{0.5cm} a>0 $$ is Laplace's integral: $$ \int_0^{\infty} e^{-\pi at^2-\pi b/t^2} \, dt = \frac{1}{2\sqrt{a}}e^{-2\pi \sqrt{ab}}, \hspace{0.5cm} a,b>0. $$ Given that the Gaussian integral may be generalized to $$ \int_0^{\infty} \cos(2\pi xt) e^{-\pi at^2} \, dt = \frac{1}{2\sqrt{a}}e^{-\pi x^2/a}, \hspace{0.5cm} a>0, \, x \in \mathbb{R}, $$ I am wondering if a similar 'cosine generalization' of Laplace's integral exists. That is, can we evaluate $$ \int_0^{\infty} \cos(2\pi xt)e^{-\pi at^2-\pi b/t^2} \, dt, \hspace{0.5cm} a,b>0, \, x \in \mathbb{R} $$ in closed form? I have tried to do this by Taylor expanding the cosine and evaluating $$ \int_0^{\infty} t^{2n} e^{-\pi at^2-\pi b/t^2} \, dt $$ for all $n \geq 0$ by differentiating Laplace's integral repeatedly with respect to $a$, but the formulas get complicated very quickly. Is there a nicer way to attempt this?

Dave
  • 1,723
  • Laplace's integral is equivalent to the Gaussian integral by Glasser's master theorem. This nice symmetry is broken by the argument of the cosine being distinct from the argument of the exponential. Because of this, I doubt there's a clean solution. – Robert Lee Dec 22 '23 at 07:21
  • However, by the same token, if you substitute the argument of the cosine by doing $t \to t - \sqrt{\frac{b}{a}}\frac{1}{t}$ then you again get a nice solution by Glasser's. – Robert Lee Dec 22 '23 at 07:23

0 Answers0