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During my studies I came across the following integral which is solvable by Mathematica. I tried to solve it by hand, but have not yet found a trick to get the desired result. To make it even worse, one try gave me a result that has to be wrong (assuming Mathematica to be correct), but I cant spot my mistake.

The integral is

$$\int_0^\infty t \exp \left({-a t -\frac{a}{c} \mathrm{e}^{-ct}} \right)\mathrm{d}t $$

Using Mathematica to give the 2-2-Hypergeometric function and expanding ${}_2F_2$ according to wikipedia: \begin{align} \int_0^\infty t \mathrm{e}^{-a t -\frac{a}{c} \mathrm{e}^{-ct}} \mathrm{d}t & = {}_2F_2 \left(\left\lbrace\frac{a}{c},\frac{a}{c}\right\rbrace;\left\lbrace\frac{a}{c}+1,\frac{a}{c}+1\right\rbrace;-\frac{a}{c} \right) \frac{1}{a^2}\\ &= \frac{1}{a^2} \sum^\infty_{k=0} \frac{\Gamma(k+a/c)\Gamma(k+a/c)}{\Gamma(a/c)\Gamma(a/c)} \frac{\Gamma(1+a/c)\Gamma(1+a/c)}{\Gamma(k+1+a/c)\Gamma(k+1+a/c)} \frac{\left(-a/c\right)^k}{k!}\\ &= \frac{1}{a^2} \sum^\infty_{k=0} \frac{(a/c)^2}{(k+a/c)^2} \frac{\left(-a/c\right)^k}{k!} \\ &= \frac{1}{c^2} \sum^\infty_{k=0} \frac{1}{(k+a/c)^2} \frac{\left(-a/c\right)^k}{k!}\\ &= \frac{1}{c^2} \sum^\infty_{k=0} g_{Mathematica}(k,a,c) \frac{\left(-a/c\right)^k}{k!} \end{align} where $a$ and $c$ are constants $>0$ and i defined the function $g_{Mathematica}(k,a,c)$.

Seeing a series pops up in the result I decided to expand the exponential function as well. Using $b=a/c$ and the bionmial expansion gave me

\begin{align} \int_0^\infty t \mathrm{e}^{-a t -b \mathrm{e}^{-c t}} \mathrm{d}t &= \int_0^\infty t \sum^\infty_{k=0} \frac{\left( -a t -b \mathrm{e}^{-c t} \right)^k}{k!} \mathrm{d} t\\ &=\int_0^\infty t \sum^\infty_{k=0} \sum^k_{l=0} \frac{k!}{l! (k-l)!} \frac{(-a t)^{k-l} (-b \mathrm{e}^{-c t})^l}{k!} \mathrm{d} t\\ &= \sum^\infty_{k=0} \sum^k_{l=0} \frac{(-a)^{k-l} (-b)^l}{l! (k-l)!} \int_0^\infty t^{k-l+1} \mathrm{e} ^{-clt} \mathrm{d} t\\ &= \sum^\infty_{k=0} \sum^k_{l=0} \frac{(-a)^{k-l} (-b)^l}{l! (k-l)!} \frac{1}{(cl)^{k-l+1}}\int_0^\infty \tau^{k-l+1} \mathrm{e}^{-\tau} \frac{\mathrm{d} \tau}{cl}\\ &= \sum^\infty_{k=0} \sum^k_{l=0} \frac{(-a)^{k-l} (-b)^l}{l! (k-l)!} \frac{1}{(cl)^{k-l+2}} \Gamma(k-l+2)\end{align} Using $b=a/c$ \begin{align} \sum^\infty_{k=0} \sum^k_{l=0} \frac{(-a)^{k-l} (-b)^l}{l! (k-l)!} \frac{(k-l+1)!}{(cl)^{k-l+2}} &= \sum^\infty_{k=0} \sum^k_{l=0} \frac{(-a)^{k-l} (-b)^l}{(cl)^{k-l+2}} \frac{(k-l+1)}{l!}\\ &= \sum^\infty_{k=0} \sum^k_{l=0} \frac{(-a)^{k-l} (-a/c)^l}{(cl)^{k-l+2}} \frac{(k-l+1)}{l!}\\ &= \frac{1}{c^2}\sum^\infty_{k=0} \sum^k_{l=0} \frac{1}{l^{k-l+2}} \frac{k-l+1}{l!} \left(-\frac{a}{c}\right)^k\\ &= \frac{1}{c^2}\sum^\infty_{k=0} g_{me}(k) \frac{\left(-a/c\right)^k}{k!}\\ \end{align}

Now, not only does this not give me what I want ($g_{me} \neq g_{Mathematica}(k,a,c)$), but also the constants $a$ and $c$ vanish in $g_{me}(k)$. Where did they go?

Any help would be appreciated!

Thanks, Best

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    this integral leads to a hypergeometric sieries – 2017-02-27
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    @Dr.SonnhardGraubner. It seems that this is written at the very start of the post. – 2017-02-27
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    yes it seems so – 2017-02-27
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    For second method, there are divergent terms in your series, namely $l=0$ and $k=0$ at first sight, you cannot intervert symbols $\int$ and $\sum$ like this. – 2017-02-27
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    @ Dr.SonnhardGraubner Thank you for your input. Do you know how to get there by hand though? @zwim could you tell me where exactly? I do not seem to find any division by zero – 2017-02-27
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    Well in $1/cl$ for instance... Also when $k=0$ then $\int_0^{\infty}t\,dt$ diverges. – 2017-02-27
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    Ah, i see. and wow yeah, i didnt think about the k=0 case. Is there anyhting i can do to still solve this integral or at least, for now, get around the divergences when expanding the exponential fct? thanks! – 2017-02-27

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