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I'm trying to find the series representation of $ f(x)=\int_{0}^{x} \frac{e^{t}}{1+t}dt $. 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 $e^{x}=\sum_{0}^{\infty}\frac{x^{n}}{n!}$ and $\frac{1}{1+x}=\sum_{}^{\infty}(-1)^{n}x^{n}$ but I don't know how to use it here. Thanks

(Don't hesitate to correct my English)

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$$e^x=\sum_{n\ge 0}\frac{x^n}{n!}$$

$$\frac{1}{x+1}=\sum_{n\ge 0}(-1)^nx^n$$

$$\frac{e^x}{x+1}=\sum_{b\ge 0}\sum_{a\ge 0} \frac{(-1)^b}{a!}x^{a+b}$$

$$\frac{e^x}{x+1}=\sum_{n\ge 0}\sum_{a+b=n}_{a,b\ge 0} \frac{(-1)^b}{a!}x^{a+b}$$

$$\frac{e^x}{x+1}=\sum_{n=0}^\infty x^n(-1)^n(\sum_{k=0}^n (-1)^{k}\frac{1}{k!})$$


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:

$$\sum_{k=0}^n\frac{(-1)^k}{k!}=\frac{1}{e}+\frac{1}{n!}(\frac{(-1)^n+1}{2}-\{ \frac{n!}{e}\})$$

Where {.} is the fractional part function.

So that we get:

$$\frac{e^x}{x+1}=\sum_{n=0}^\infty x^n(-1)^n(\frac{1}{e}+\frac{1}{n!}(\frac{(-1)^n+1}{2}-\{ \frac{n!}{e}\}))$$

$$=\frac{1}{e}\sum_{n=0}^\infty x^n(-1)^n+\sum_{n=0}^\infty x^n(-1)^n\frac{1}{n!}(\frac{(-1)^n+1}{2}-\{ \frac{n!}{e}\})$$

$$=\frac{1}{e(x+1)}+\frac{1}{2}\sum_{n=0}^\infty\frac{x^n}{n!}(1+(-1)^n)-\sum_{n=0}^\infty\frac{x^n(-1)^n}{n!}\{ \frac{n!}{e}\}$$ $$=\frac{1}{e(x+1)}+\frac{e^x+e^{-x}}{2}-\sum_{n=0}^\infty\frac{x^n(-1)^n}{n!}\{ \frac{n!}{e}\}$$ $$=\frac{1}{e(x+1)}+\frac{e^x+e^{-x}}{2}+\sum_{n=0}^\infty\frac{x^n}{n!}\{ \frac{n!}{e}\}(-1)^{n+1}$$

Now integrating term by term gives,

$$\int_{0}^x \frac{e^t}{1+t} dt=\frac{\ln(x+1)}{e}+\frac{e^x-e^{-x}}{2}+\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}\{ \frac{n!}{e}\}(-1)^{n+1}$$


Thus for all $|x|<1$ the following expansion holds:

$$f(x)=\int_{0}^x \frac{e^t}{1+t} dt=\frac{\ln(x+1)}{e}+\frac{e^x-e^{-x}}{2}+\sum_{n=1}^\infty\frac{(-x)^n}{n!}\{ \frac{(n-1)!}{e}\}$$

Also pretty off topic, but the positive $n^{\text{th}}$ 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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A hint: if $$g(t) = \sum_{n=0}^\infty a_nt^n$$ and $$h(t) = \sum_{n=0}^\infty b_nt^n$$ then $$g(t)h(t) = \sum_{n=0}^\infty\left(\sum_{k=0}^n a_kb_{n-k}\right)t^n.$$

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