All Questions
86
questions
4
votes
1
answer
91
views
Why $\infty=\sum_{i=1}^\infty \frac{1}{n+i}\neq\lim_{n\rightarrow\infty}\sum_{i=1}^n \frac{1}{n+i}=\log2$?
I was wondering why $\sum_{i=1}^\infty \frac{1}{n+i}$ diverges but $\lim_{n\rightarrow\infty}\sum_{i=1}^n \frac{1}{n+i}=\log2$. While assuming integral as limit of series, we find out that:
$$
\int_1^...
- 43
4
votes
1
answer
168
views
Prove that sum of integrals $= n$ for argument $n \in \mathbb{N}_{>1}$
ORIGINAL QUESTION (UPDATED):
I have a function $f:\mathbb{R} \rightarrow \mathbb{R}$ containing an integral that involves the floor function:
$$f(x):= - \lfloor x \rfloor \int_1^x \lfloor t \rfloor x \...
2
votes
0
answers
364
views
A sum of two curious alternating binoharmonic series
Happy New Year 2024 Romania!
Here is a question proposed by Cornel Ioan Valean,
$$\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{2^{2n}}\binom{2n}{n}\sum_{k=1}^n (-1)^{k-1}\frac{H_k}{k}-\sum_{n=1}^{\infty}(-1)...
- 5,495
0
votes
0
answers
63
views
Converting complex-exponential summation to Fresnel integrals
I have a summation
$$S = \sum_{n=0}^N e^{-jn^2a}, \ a\ne 0, \ n\in\{0,1,\cdots,N\}$$
and it can be approximated by
$$S\approx I = \int_{n=0}^N e^{-jn^2a}dn$$
when $N$ is sufficiently large. ...
- 71
0
votes
1
answer
80
views
Prove that $\int_0^1\lfloor nx\rfloor^2 dx = \frac{1}{n}\sum_{k=1}^{n-1} k^2$
First of all apologies for the typo I made in an earlier question, I decided to delete that post and reformulate it
I am asked to prove that
$$\int_{(0,1)} \lfloor nx\rfloor^2\,\mathrm{d}x =\frac{1}{n}...
- 131
3
votes
2
answers
180
views
Interchanging integration with summation on a specific example
Consider the infinite sum
$$S=\sum_{j=1}^\infty (-1)^{j +1}\frac{1}{2j+1} \int_0^1 \frac{1-x^{2j}}{1-x}dx$$
and the associated integral
$$L=\int_0^1 \sum_{j=1}^\infty\frac{ 1-x^{2j}}{1-x} (-1)^{j +1}\...
- 482
2
votes
1
answer
112
views
Complete the proof of $ \int_{-\infty}^{\infty}{e^{-x^2/2}x^{2n} dx} = \sqrt{2\pi} \frac{(2n)!}{2^nn!} $
What I already know is that if $L(s) := \int_{-\infty}^{\infty}{e^{sx}e^{\frac{-x^2}{2}} dx}=\sqrt{2\pi}e^{\frac{s^2}{2}}$, then $L^{(2n)}(0)=\int_{-\infty}^{\infty}{e^{\frac{-x^2}{2}}x^{2n} dx}$. So ...
- 65
1
vote
1
answer
74
views
Estimating a sum with an integral to nearest power of $10$
Suppose I want to estimate the sum $\sum_{x = -1000}^{1000} \sum_{y: x^2 + y^2 < 10^6 } x^2$, or identically, $\sum_{y = -1000}^{1000} \sum_{x: x^2 + y^2 < 10^6 } x^2$, to the nearest power of $...
- 737
1
vote
1
answer
52
views
A Riemann Sum clarification
I have to find the area under the curve $f(x) = x^2$ in the segment $[-2, 1]$ using Riemann Sum. So here is what I did, but it's wrong and I need some clarification about (see the questions in the end)...
- 3,482
2
votes
1
answer
59
views
Integration with Riemann Sum
I wanted to try to perform a Riemann Sum for the following integral, but I got stuck in the middle.
$$\int_{-1}^0 e^{-x^2}\ \text{d}x$$
So the interval is $[-1, 0]$, and I chose $\Delta x = \dfrac{1}{...
- 3,482
0
votes
0
answers
46
views
Why can we approximate a sum by a definite integral?
From wikipedia https://en.wikipedia.org/wiki/Summation#Approximation_by_definite_integrals, I read that
$\int_{s=a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{s=a}^{b+1} f(s)\ ds$ for increasing ...
- 365
1
vote
1
answer
97
views
How to evaluate $\sum_{k=0}^{\infty} \frac{(-1)^{k}(2k+1)!!}{(k+1)(2k+2)!!}\frac{\pi}{2}\alpha^{k+1}$
I was trying to solve the integral $\int_0^{\frac{\pi}{2}} \ln(1+\alpha\sin^2 x)\, dx$, where $\alpha>-1$
And my approach is using infinite sum by expanding $\ln(1+\alpha\sin^2 x)$
So let $I=\int_0^...
- 858
4
votes
2
answers
201
views
Evaluate the following humongous expression
PROBLEM:
Evaluate $$\left(\frac{\displaystyle\sum_{n=-\infty}^{\infty}\frac{1}{1+n^2}}{\operatorname{coth}(\pi)}\right)^2$$
CONTEXT:
I saw a very interesting and yet intimidating question on the ...
- 1,253
0
votes
0
answers
56
views
prove that $\sum_{i,j=1}^n f(a_i - a_j) = \int_{-\infty}^\infty (\sum_{i=1}^n \frac{1}{1+(x-a_i)^2})^2 dx,$
Prove that if $a_1,a_2,\cdots, a_n$ are real numbers then $$\sum_{i,j=1}^n f(a_i - a_j) = \int_{-\infty}^\infty \left(\sum_{i=1}^n \frac{1}{1+(x-a_i)^2}\right)^2 dx,$$ where $f(y) = \int_{-\infty}^\...
- 1,137
1
vote
0
answers
71
views
Evaluate $\int_1^e\frac{1}{x}dx$ using upper and lower sums
Using the upper and lower sums, I've tried to solve $\int_1^e\frac{1}{x}dx$ in the following way.
Let $$\Delta x = \frac{\mathit e -1}{n}$$ and let a partition $P$ be given by $$ P = \{1,1+\Delta x, 1+...
- 11