Is it possible to find the value of $\sum_{n = 0}^\infty n \int_{n}^{n+1}e^{-x^2}dx$?

Asked May 08 '22 at 13:15

Active May 08 '22 at 13:18

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I want to calculate $$ \sum_{n = 0}^\infty n \int_{n}^{n+1}e^{-x^2}dx. $$

Since $$ \sum_{n = 0}^\infty n \int_{n}^{n+1}e^{-x^2}dx \leq \sum_{n = 0}^\infty \int_{n}^{n+1}xe^{-x^2}dx = \int_0^\infty xe^{-x^2} dx $$ and $\int_0^\infty xe^{x^2}dx = 1/2$, it is followed that the sum converges to some value. But I have a trouble to find this value.

What I tried:

Since $e^{-(n+1)^2} \leq \int_n^{n+1} e^{-x^2} dx \leq e^{-n^2}$, $$ \sum_{n = 0}^\infty ne^{-(n+1)^2} \leq \sum_{n = 0}^\infty n \int_{n}^{n+1}e^{-x^2}dx \leq \sum_{n = 0}^\infty ne^{-n^2} $$

Please some help or advice. Thank you.

edited May 08 '22 at 13:18

Gary

31,845

asked May 08 '22 at 13:15

jamize

3

Your value of $0.1619…$ looks like this question. Possible duplicate? – Тyma Gaidash May 08 '22 at 13:19
It can be written $\int_0^\infty \lfloor x\rfloor ,e^{-x^2},\mathrm d x$. But why do you think it can be simplified? – LL 3.14 May 08 '22 at 13:21
Yes, this is what I was looking for. I did not do enough research. My apologies and thank you ! – jamize May 08 '22 at 13:31
This is an interesting question. I've closed it as a duplicate of another (interesting) question. But I note that there is a difference of exactly $1$ between the answer of this question and that question. – davidlowryduda Feb 14 '24 at 20:03

Is it possible to find the value of $\sum_{n = 0}^\infty n \int_{n}^{n+1}e^{-x^2}dx$?

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