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Polynomials being a subspace of Banach Space $(C[-1,1],\lVert \cdot \rVert_{\infty})$, must have empty interior, otherwise it'd have to be the full space. (One argument for showing it's not closed, Polynomials have a countable Hamel basis and Banach spaces of infinite dimension cannot have a countab...

Since, $T$ is Hausdorff, the graph of the continous linear map $i \circ f$ is closed subspace of $V \times T$. Again, the linear map, ${Id}_{V} \times i : V \times W \to V \times T$ is continuous and $({Id}_{V} \times i )^{-1} (G_{i \circ f}) = G_f$ (since, $i$ is a injective map). \begin{align*}G_{...

To justify that the integral does not converge, we note that if $\displaystyle F(A) = \int_0^{A} f(x)\,dx$ converge as $A \to \infty$, then $F(A+\alpha) - F(A) \to 0$ as $A \to \infty$ (for any $\alpha >0$). But, this is not the case here, \begin{align*}\int_{2k\pi}^{(2k+1)\pi} \sin \left(\sqrt{x}\r...

The integral is oscillating at infinity (does not converge). However, we can still define it in some limiting sense by taking $a \to 0$ here http://www.mathimatikoi.org/forum/viewt ... p=902#p902

Hey T, thanks for sharing my problem! :) Here's the original motivation behind creation of the problem: Consider, the partial fraction decomposition of, $\displaystyle f_n(x) = \frac{x^{m}}{\prod\limits_{k=1}^{n}(x+k)} = \sum\limits_{k=1}^{n} \frac{A_k}{x+k}$ for some $m < n$, then for each $1 \le k...

Making the change in variable $x \mapsto -ex$, the problem is equivalent to the determination of: \begin{align*} \lim_{x\to-e^+}\int_{0}^{+\infty}\frac{(x+e)^{1/2}}{e^t+xt}\,dt = \lim_{x \to 1^{-}} \frac{(1-x)^{1/2}}{\sqrt{e}}\int_{0}^{\infty}\frac{1}{e^{t-1} - xt}\,dt\end{align*} Let, $x \in (0,1)$...

The number of ways of writing an integer as sum of two square integers (both positive and negative), $\displaystyle r_2(n) = 4\sum\limits_{\substack{d |n\\ d \text{ odd }}} (-1)^{\frac{d-1}{2}}$. Then, for $\mathfrak{Re}(s) > 1$, we have: \begin{align*}\sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus ...

\begin{align*}\sum_{n=0}^{\infty} \left(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{n+k}\right)^2 &= \sum\limits_{n=0}^{\infty} \left((-1)^{n}\int_0^{1} \frac{x^n}{1+x}\,dx\right)^2\\&= \sum\limits_{n=0}^{\infty} \int_0^{1}\int_0^1 \frac{x^ny^n}{(1+x)(1+y)}\,dx\, dy\\&= \int_0^{1} \int_0^1 \frac...

Taking logarithm on both sides, \begin{align*}&\log x + \sum\limits_{n=1}^{\infty}\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)} \log \left(\frac{n+x}{n+1}\right) \\&= \sum\limits_{n=0}^{\infty} \frac{1}{4^n}\binom{2n}{n} \int_{0}^{\infty} \frac{e^{-(n+1)t} - e^{-(n+x)t}}{t}\,dt\\...

\begin{align}\int_0^1 \frac{x\log x}{\sqrt{1-x^2}} &= \frac{1}{4}\int_0^1 \frac{\log y}{\sqrt{1-y}}\,dy\\&= \frac{1}{2}\int_0^1 \log (1-z^2)\,dz\\&= \frac{1}{2}[z\log (z) - z + (z+1)\log (z+1) - (z+1)]_0^1 \\&= \log 2 - 1\end{align} Made the change of variable, $y = x^2$ in step $(1)...