I'm trying to find the series representation of f(x)=∫x0et1+tdt. I have found it using the Maclaurin series, differentiating multiple times and finding a pattern. But I think there must be an easier way, using the power series of elementary functions. I know that ex=∑∞0xnn! and 11+x=∑∞(−1)nxn but I don't know how to use it here. Thanks
(Don't hesitate to correct my English)
Answer
ex=∑n≥0xnn!
1x+1=∑n≥0(−1)nxn
exx+1=∑b≥0∑a≥0(−1)ba!xa+b
\frac{e^x}{x+1}=\sum_{n\ge 0}\sum_{a+b=n}_{a,b\ge 0} \frac{(-1)^b}{a!}x^{a+b}
exx+1=∞∑n=0xn(−1)n(n∑k=0(−1)k1k!)
Now the interesting thing about these coefficients above that are given in the form of partial sums is that it turns out that we have:
n∑k=0(−1)kk!=1e+1n!((−1)n+12−{n!e})
Where {.} is the fractional part function.
So that we get:
exx+1=∞∑n=0xn(−1)n(1e+1n!((−1)n+12−{n!e}))
=1e∞∑n=0xn(−1)n+∞∑n=0xn(−1)n1n!((−1)n+12−{n!e})
=1e(x+1)+12∞∑n=0xnn!(1+(−1)n)−∞∑n=0xn(−1)nn!{n!e}
=1e(x+1)+ex+e−x2−∞∑n=0xn(−1)nn!{n!e}
=1e(x+1)+ex+e−x2+∞∑n=0xnn!{n!e}(−1)n+1
Now integrating term by term gives,
∫x0et1+tdt=ln(x+1)e+ex−e−x2+∞∑n=0xn+1(n+1)!{n!e}(−1)n+1
Thus for all |x|<1 the following expansion holds:
f(x)=∫x0et1+tdt=ln(x+1)e+ex−e−x2+∞∑n=1(−x)nn!{(n−1)!e}
Also pretty off topic, but the positive nth coefficients that took the form of partial sums in our first list of series manipulations above correspond to solutions to the hat-check problem when n hats are being considered.
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