Calculate the integral $\int_{0}^{\infty} t^{-3/2}e^{-a/bt-ct}\,\mathrm{d}t$

Question

I am trying to calculate the following integral: $$\int_{0}^{\infty} t^{-3/2}e^{-a/(bt)-ct}\,\mathrm{d}t,$$ where $a,b,c$ are real constants.

I tried to use a similar approach as Gamma function but I did not manage to get anything.

I do not want the solution, just a hint for solving it.

Thank you.

EDIT: Sorry for the misunderstanding. I corrected the notation.

Answer 1

$$\int_{0}^{\infty} t^{-3/2}e^{-a/bt-ct}\,\mathrm{d}t$$

where $f(t)=t^{-3/2}e^{-a/bt-ct}=t^{-3/2}e^{-(a/b+c)t}$

take $u=(a/b+c)t$

$$\sqrt{a/b+c}\int_{0}^{\infty} (a/b+c)^{-3/2}t^{-3/2}(a/b+c)e^{-(a/b+c)t}\,\mathrm{d}t$$

$$\sqrt{a/b+c}\int_{0}^{\infty} u^{-3/2}e^{-u}\mathrm{d}u$$ $$\sqrt{a/b+c} \times \Gamma(5/2)$$ $$\sqrt{a/b+c}\times \frac{3}{2}\times \frac{\sqrt{\pi}}{2}$$

Is this what you were looking for? I wanted to give a hint, but then the hint would've revealed the final solution directly anyway.

2nd Interpretation:

$$\int_{0}^{\infty} f(t) \mathrm{d}t$$

where $f(t)=t^{-3/2}\exp({-\frac{a}{bt}-ct})$

Using 10.32.10 in here, take the following substitutions:

$$u=ct$$ $$z=2\sqrt{\frac{ac}{b}}$$ $$v=\frac{1}{2}$$

Then we will have:

$$\frac{c}{2 \times ({\frac{ac}{b}})^{1/4}} \times K_{1/2}(2\sqrt{\frac{ac}{b}}) = \int_{0}^{\infty} f(u) \mathrm{d}u$$