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?


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}]
  • $\begingroup$ 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

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 $$ \frac{1}{a^2}\sum_{m=1}^{n}\sinh\left(\frac{a^2}{m}\right).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.