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EDITED.

I am trying to find a closed form for this integral:

$$\int_0^{\infty } \frac{e^{-n t} \left(-1+e^{n t}\right) (\, _0F_1(;2;a t)+\, _0F_1(;2;b t))}{2 \left(-1+e^t\right)} \, dt $$ where:$\, _0F_1(;2;a t)$ and $\, _0F_1(;2;b t)$ is the confluent hypergeometric function.

$a$,$b$, $n$, are positive constants, and $n>1$, $n\in \mathbb{Z}$

Does anyone have any suggestions or can advise?

Thanks

MMA code:

 Integrate[(E^(-n t) (-1 + E^(n t)) (Hypergeometric0F1[2, a t] + 
 Hypergeometric0F1[2, b t]))/(2 (-1 + E^t)), {t, 0, Infinity}]
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  • $\begingroup$ Have you attempted the "brutal approach" of expanding $I_1+J_1$ as its Taylor series, $\frac{e^{-nt^2}-1}{e^{t^2}-1}$ as a geometric series and exploiting the fact that $\int_{0}^{+\infty}t^m e^{-nt^2}\,dt$ is known? $\endgroup$ – Jack D'Aurizio Jul 12 '17 at 15:47
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For any $a>0$ and $n\in\mathbb{N}^*$ we have $$ \int_{0}^{+\infty}\left[I_1(2at)+J_1(2at)\right]e^{-nt^2}\,dt = \frac{1}{a}\sinh\left(\frac{a^2}{n}\right)$$ hence the given integral depends on $$ \frac{1}{a^2}\sum_{m=1}^{n}\sinh\left(\frac{a^2}{m}\right).$$

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