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I'm trying to evaluate the following integral which popped up in MIT Integration Bee 2015 which involves the floor function.

$$\int_{0}^{\infty}\left(xe^{1-x}-\lfloor x\rfloor e^{1-\lfloor x\rfloor}\right)\mathrm dx$$

My Attempt:

$$\begin{aligned}\mathrm I &=\int_{0}^{\infty}xe^{1-x}\mathrm dx-\sum_{n=0}^{\infty}\int_{n}^{n+1}ne^{1-n}\mathrm dx\\ &= e-\sum_{n=0}^{\infty}ne^{1-n}=e\biggl(1+\sum_{n=0}^{\infty}\left(e^{1-n}\right)'\biggr)\end{aligned}$$


I'm getting stuck at this step because I'm not able to figure out how to evaluate this sum. I know one way is to use differentiation to get an expression for the sum but I'm not sure how to proceed. A hint in the right direction would be appreciated. Thanks

Note: This is different from How can I evaluate $\sum_{0}^{\infty}(n+1)x^n$? . This problem is about bringing the sum into the form from which differentiation would yield the result.

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    $\begingroup$ Note that$$\sum_{n=0}^\infty ne^{1 - n} = \sum_{m=-1}^\infty (m + 1)e^{-m} = \sum_{m=0}^\infty (m + 1)(e^{-1})^m.$$ $\endgroup$ – Theo Bendit May 15 at 6:06
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    $\begingroup$ You can't differentiate with respect to $n$, Paras. But you can differentiate $\sum_ne^{1-n}x^n$ with respect to $x$. $\endgroup$ – Gerry Myerson May 15 at 6:23
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    $\begingroup$ Precisely and then setting $x=1$ gives the result. Thanks @GerryMyerson :) $\endgroup$ – Paras Khosla May 15 at 6:27
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    $\begingroup$ Ah. Why did this not occur to me. Thanks @TheoBendit :) $\endgroup$ – Paras Khosla May 15 at 6:29
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    $\begingroup$ So, if you can do the problem now, Paras, let me encourage you to write up and post an answer. $\endgroup$ – Gerry Myerson May 15 at 6:31
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Thanks to motivation from @Gerry Myerson. I'm attempting to answer my own question. Let the integral in question be denoted by $\mathrm I$ .$$\begin{aligned}\mathrm I &=\int_{0}^{\infty}xe^{1-x}\mathrm dx-\sum_{n=0}^{\infty}\int_{n}^{n+1}ne^{1-n}\mathrm dx\\ &= e-\sum_{n=0}^{\infty}ne^{1-n}=e-\sum_{n=0}^{\infty}\left(e^{1-n}x^n\right)'\end{aligned}$$

For the sum, differentiating $\sum_{n=0}^{\infty}e^{1-n}x^n$ term-by-term is the way to go. $$e\sum_{n=0}^{\infty}\left(\dfrac{x}{e}\right)^n=\dfrac{e^2}{e-x}\implies \sum_{n=0}^{\infty}ne^{1-n}x^{n-1}=\dfrac{e^2}{(e-x)^2}$$

Plugging in $x=1$ and consequently the value of the infinite sum into the value of the expression for the integral yields: $$\mathrm I =e-\dfrac{e^2}{(e-1)^2}=\dfrac{e^3-3e^2+e}{e^2-2e+1}$$

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