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Does $$\int_0^a e^{-bt-c/t} dt \tag{1}$$ have a closed-form expression?

Note: $a, b,$ and $c$ are all positive constants.

The following is my trials.

(i) Let $x = -bt-\frac{c}{t}$. But I can't express $t$ in terms of $x$.

(ii) Let $y = \frac{1}{t}$. Then $t = \frac{1}{y}$, $\frac{dy}{dt} = - \frac{1}{t^2} = -y^2 \Rightarrow dt = - \frac{dy}{y^2}$, and $t = 0 \Rightarrow y=\frac{1}{0}$. But zero cannot appear in the denominator.

Then I don't know how to continue.

  • 4
    My guess would be no – 2017-02-14
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    if you let the upper boundary be infinity, then yes – 2017-02-14
  • 2
    @tired. The modified Bessel function of the second kind ? Do I (at my age) remember properly ? – 2017-02-14
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    @JanEerland. Don't you think that there is a problem at $t=0$ ? – 2017-02-14
  • 1
    [Don't use `\frac` in exponents or limits of integrals](https://math.meta.stackexchange.com/a/5057/290189). – 2018-03-15
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    By infinite sum : $\int_0^a e^{-\frac{c}{t}-b t} \, dt=\sum _{j=0}^{\infty } \frac{(-b)^j c^{1+j} \Gamma \left(-1-j,\frac{c}{a}\right)}{j!}$ – 2018-03-15

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The integral can be written as \begin{align} \int_0^a e^{-bt-c/t} \,dt&=a\int_0^1 e^{-abt-\tfrac{c}{at}}\,dt\\ &=a\int_1^\infty e^{-cu/a-ab/u}\,\frac{du}{u^2}\\ &=aK_1\left( \frac{c}{a},ab \right) \end{align} where the integral corresponds to the definition of the incomplete Bessel function by Harris here or here also known as the leaky aquifer function. You may find useful information in papers by Jones (Proc. Edimb. Math. Soc. 2007) for its asymptotic expansions, Fripiat, Harris and many others for its general properties and numerical evaluation.