Solving a Triple Infinite Series with an Exponential Integral
Published July 9, 2026, 7:46 p.m. by james
The value is:
$$ \boxed{\frac{94}{375}} $$
We have
$$ S = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \int_0^\infty \frac{x^{a+b+c}}{(a+b+c)!} e^{-4x}\cosh(2x)\,dx $$
Step 1: Combine the Three Indices
Let
$$ n=a+b+c. $$
For each $n \ge 3$, the number of positive integer triples $(a,b,c)$ satisfying $a+b+c=n$ is
$$ \binom{n-1}{2}. $$
Therefore,
$$ S = \sum_{n=3}^{\infty} \binom{n-1}{2} \int_0^\infty \frac{x^n}{n!} e^{-4x}\cosh(2x)\,dx. $$
Step 2: Simplify the Exponential Term
Using
$$ \cosh(2x)=\frac{e^{2x}+e^{-2x}}{2}, $$
we get
$$ e^{-4x}\cosh(2x) = \frac{e^{-2x}+e^{-6x}}{2}. $$
Step 3: Evaluate the Integral
Thus,
$$ \int_0^\infty \frac{x^n}{n!} e^{-4x}\cosh(2x)\,dx = \frac{1}{2n!} \left( \int_0^\infty x^n e^{-2x}\,dx + \int_0^\infty x^n e^{-6x}\,dx \right). $$
Using
$$ \int_0^\infty x^n e^{-\lambda x}\,dx = \frac{n!}{\lambda^{n+1}}, $$
we get
$$ \int_0^\infty \frac{x^n}{n!} e^{-4x}\cosh(2x)\,dx = \frac12 \left( \frac{1}{2^{n+1}} + \frac{1}{6^{n+1}} \right). $$
Step 4: Sum the Series
Therefore,
$$ S = \frac12 \sum_{n=3}^{\infty} \binom{n-1}{2} \left( \frac{1}{2^{n+1}} + \frac{1}{6^{n+1}} \right). $$
Use the identity
$$ \sum_{n=3}^{\infty} \binom{n-1}{2}t^{n+1} = \frac{t^4}{(1-t)^3}. $$
Substituting $t=\frac12$ and $t=\frac16$, we get
$$ S = \frac12 \left[ \frac{(1/2)^4}{(1-1/2)^3} + \frac{(1/6)^4}{(1-1/6)^3} \right]. $$
Now simplify:
$$ S = \frac12 \left[ \frac12+\frac{1}{750} \right]. $$
Hence,
$$ S = \frac14+\frac{1}{1500} = \frac{375}{1500} + \frac{1}{1500} = \frac{376}{1500} = \frac{94}{375}. $$
Therefore,
$$ \boxed{\frac{94}{375}} $$
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