Laplace transformation of a polynomial function involving square root (or approximation of the integral)

Question

Could somebody suggest how to solve this Laplace transform: $$ \int_0^\infty{e^{-at}\over\sqrt{A+Bt+Ct^2}}{\rm d\,}t $$ ?

The real coefficients $A,B,C$ are chosen such that the roots of $A+Bt+Ct^2$ lie on the negative axis. The approach from here Laplace transform of $ t^{1/2}$ and $ t^{-1/2}$ does not seem to work.

My try: I tried to multiply the integrand by $A+Bt+Ct^2$ leading to $$ \int_0^\infty{e^{-at}\sqrt{A+Bt+Ct^2}}\ {\rm d\,}t. $$ If this can be solved then by using the identity $$ \mathcal{L}[t^nf(t)]=(-1)^n F^{(n)}(a), $$ where $$ \mathcal{L}[f(t)]=F(a) $$ ( $f(t)=1/\sqrt{A+Bt+Ct^2}$ ) I could integrate the right side and get the result. But I can't solve the second integral either (per partes did not work).

EDIT: If no expression exists, is there a way to find an asymptotic form of the integral $$ \int_0^K{e^{-at}\over\sqrt{A+Bt+Ct^2}}{\rm d\,}t $$ for fixed $a$ and $K\to\infty$ ?

Answer 1

$\int_0^\infty\dfrac{e^{-at}}{\sqrt{A+Bt+Ct^2}}dt$

$=\int_0^\infty\dfrac{e^{-at}}{\sqrt{C\left(t^2+\dfrac{Bt}{C}+\dfrac{A}{C}\right)}}dt$

$=\int_0^\infty\dfrac{e^{-at}}{\sqrt{C\biggl(t^2+\dfrac{Bt}{C}+\dfrac{B^2}{4C^2}+\dfrac{A}{C}-\dfrac{B^2}{4C^2}\biggr)}}dt$

$=\int_0^\infty\dfrac{e^{-at}}{\sqrt{C\biggl(t+\dfrac{B}{2C}\biggr)^2-\dfrac{B^2-4AC}{4C}}}dt$

$=e^\frac{aB}{2C}\int_\frac{B}{2C}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$

Case $1$: $C>0$ and $B^2-4AC>0$

Then $e^\frac{aB}{2C}\int_\frac{B}{2C}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$

$=e^\frac{aB}{2C}\int_\frac{B}{2C}^\sqrt\frac{B^2-4AC}{4C^2}\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt+e^\frac{aB}{2C}\int_\sqrt\frac{B^2-4AC}{4C^2}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$

For $\int_\sqrt\frac{B^2-4AC}{4C^2}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$ ,

$\int_\sqrt\frac{B^2-4AC}{4C^2}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$

$=\int_0^\infty\dfrac{e^{-a\sqrt{\frac{B^2-4AC}{4C^2}}\cosh t}}{\sqrt{C\left(\sqrt{\dfrac{B^2-4AC}{4C^2}}\cosh t\right)^2-\dfrac{B^2-4AC}{4C}}}d\left(\sqrt{\dfrac{B^2-4AC}{4C^2}}\cosh t\right)$

$=\int_0^\infty\dfrac{e^{-a\sqrt{\frac{B^2-4AC}{4C^2}}\cosh t}\sqrt{\dfrac{B^2-4AC}{4C^2}}\sinh t}{\sqrt{\dfrac{B^2-4AC}{4C}\cosh^2t-\dfrac{B^2-4AC}{4C}}}dt$

$=\int_0^\infty e^{-a\sqrt{\frac{B^2-4AC}{4C^2}}\cosh t}~dt$

$=K_0\left(a\sqrt{\dfrac{B^2-4AC}{4C^2}}\right)$

But for $\int_\frac{B}{2C}^\sqrt\frac{B^2-4AC}{4C^2}\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$ , it seems that this is already the simplest approach:

$\int_\frac{B}{2C}^\sqrt\frac{B^2-4AC}{4C^2}\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt=\int_\frac{B}{2C}^\sqrt\frac{B^2-4AC}{4C^2}\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nt^n}{n!\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$

Then handle $\int\dfrac{t^n}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$ , where $n$ is any non-negative integer.

Case $2$: $C>0$ and $B^2-4AC<0$

Then $e^\frac{aB}{2C}\int_\frac{B}{2C}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$

$=e^\frac{aB}{2C}\int_\frac{B}{2C}^0\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt+e^\frac{aB}{2C}\int_0^\infty\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$

For $\int_0^\infty\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$ ,

$\int_0^\infty\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$

$=\int_0^\infty\dfrac{e^{-a\sqrt{\frac{4AC-B^2}{4C^2}}\sinh t}}{\sqrt{C\left(\sqrt{\dfrac{4AC-B^2}{4C^2}}\sinh t\right)^2+\dfrac{4AC-B^2}{4C}}}d\left(\sqrt{\dfrac{4AC-B^2}{4C^2}}\sinh t\right)$

$=\int_0^\infty\dfrac{e^{-a\sqrt{\frac{4AC-B^2}{4C^2}}\sinh t}\sqrt{\dfrac{4AC-B^2}{4C^2}}\cosh t}{\sqrt{\dfrac{4AC-B^2}{4C}\sinh^2t+\dfrac{4AC-B^2}{4C}}}dt$

$=\int_0^\infty e^{-a\sqrt{\frac{4AC-B^2}{4C^2}}\sinh t}~dt$

$=\dfrac{\pi}{2}\mathbf{K}_0\left(a\sqrt{\dfrac{4AC-B^2}{4C^2}}\right)$

But for $\int_\frac{B}{2C}^0\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$ , it seems that this is already the simplest approach:

$\int_\frac{B}{2C}^0\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt=\int_\frac{B}{2C}^0\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nt^n}{n!\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$

Then handle $\int\dfrac{t^n}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$ , where $n$ is any non-negative integer.