integration - How to evaluate this improper integral $int_{0}^{infty}frac{1-x}{1-x^{n}},mathrm dx$?

$\def\d{\mathrm{d}}$How to evaluate this improper integral?
$$\int_{0}^{\infty}\frac{1-x}{1-x^{n}}\,\d x.$$

What I tried is a substitution i.e $x^{n}=t$, but then things got complicated, and I'm stuck.

Answer

Here are two methods,using real analysis

Method : 1 : using some special functions :
$$I=\displaystyle \int_{0}^{\infty}\dfrac{1-x}{1-x^{n}}\,dx=\underbrace{\displaystyle \int_{0}^{1}\dfrac{1-x}{1-x^{n}}\,dx}_{I_{1}(n)}+\underbrace{\displaystyle \int_{1}^{\infty}\dfrac{1-x}{1-x^{n}}\,dx}_{I_{2}(n)} $$
$$I_{1}(n)=\displaystyle \int_{0}^{1}\dfrac{1-x}{1-x^{n}}\,dx$$
Now lets make a substitution , $x^{n}=t \implies dx=\dfrac{1}{n}t^{\frac{1}{n}-1}dt$ , so $I_{1}(n)$ becomes
$$I_{1}(n)=\dfrac{1}{n}\displaystyle \int_{0}^{1}\dfrac{1-t^{\frac{1}{n}}}{1-t}.t^{\frac{1}{n}-1}\,dt= \dfrac{1}{n}\left[\displaystyle
\int_{0}^{1}\dfrac{t^{\frac{1}{n}-1}}{1-t}\,dt-\displaystyle \int_{0}^{1}\dfrac{{t^{\frac{2}{n}-1}}}{1-t}\,dt \right]$$
$$I_{1}(n)= \dfrac{1}{n}\displaystyle \lim_{m \rightarrow 0}\left[ \beta\left(m,\dfrac{1}{n}\right)-\beta \left(m,\dfrac{2}
{n}\right)\right] $$
$$\large \boxed{I_{1}(n)=\dfrac{1}{n} \left[\psi\left(\dfrac{2}{n}\right)-\psi \left(\dfrac{1}{n}\right)\right]}$$

now lets take $I_{2}(n)$
$$I_{2}(n)=\displaystyle \int_{1}^{\infty}\dfrac{1-x}{1-x^{n}}\,dx$$
Now for this one substitute $x=\dfrac{1}{t} \implies dx=-\dfrac{1}{t^{2}} dt$
$$I_{2}(n)=\displaystyle \int_{0}^{1}\dfrac{1-t}{1-t^{n}}t^{n-3}\,dt$$
Now make another substitution $t^{n}=u \implies dt=\dfrac{1}{n}u^{\frac{1}{n}-1}du$ , so the integral becomes
$$I_{2}(n)=\dfrac{1}{n}\displaystyle \int_{0}^{1}\dfrac{1-u^{\frac{1}{n}}}{1-u}\left(u^{\frac{1}{n}-1}\right)\left(u^{1-\frac{3}{n}}\right)\,du$$
$$I_{2}(n)= \dfrac{1}{n}\left[\displaystyle \int_{0}^{1}\dfrac{u^{-\frac{2}{n}}}{1-u}\,du-\displaystyle \int_{0}^{1}\dfrac{{u^{-\frac{1}{n}}}}{1-u}\,du \right]$$
$$I_{2}(n)= \dfrac{1}{n}\displaystyle \lim_{m \rightarrow 0}\left[ \beta\left(m,1-\dfrac{2}{n}\right)-\beta \left(m,1-\dfrac{1}{n}\right)\right]$$
$$\large \boxed{I_{2}(n)= \dfrac{1}{n}\left[ \psi\left(1-\dfrac{1}{n}\right)-\psi \left(1-\dfrac{2}{n}\right)\right]}$$
Now we just have to add $I_{1}(n)$ and $I_{2}(n)$

$$\begin{equation} I(n)=\dfrac{1}{n}\left[ -\psi\left(\dfrac{1}{n}\right)+\psi\left(1-\dfrac{1}{n}\right)+\psi\left(\dfrac{2}{n}\right)-
\psi \left(1-\dfrac{2}{n}\right)\right]\\= \dfrac{\pi}{n}\left[ \cot \left(\dfrac{\pi}{n}\right)-\cot \left(\dfrac{2\pi}{n}\right)\right]\\=
\dfrac{\pi}{n}\csc\left(\dfrac{2 \pi}{n}\right)\end{equation}$$
$$\Large \displaystyle\ \bbox[10pt, border:2pt solid #06f]{I(n)=\dfrac{\pi}{n}\csc\left(\dfrac{2 \pi}{n}\right)} \tag*{}$$

hete $\psi\Rightarrow $Digamma function (https://en.wikipedia.org/wiki/Digamma_function) and $\beta\Rightarrow $Beta function

(https://en.wikipedia.org/wiki/Beta_function#Derivatives).
and the reflection formula i used to simplify is known as Euler’s reflection formula which is …
$$\boxed{\psi(1-z)-\psi(z)=\pi \cot(\pi z)} $$

Method : 2 : Mellin Transform(http://mathworld.wolfram.com/MellinTransform.html) :
$$I(n)=\displaystyle \int_{0}^{\infty}\dfrac{1-x}{1-x^{n}}\,dx$$
so lets use the same substitution again ..
$x^{n}=t \implies dx=\dfrac{1}{n}t^{\frac{1}{n}-1}dt $ so the integral becomes
$$I(n)=\dfrac{1}{n} \displaystyle \int_{0}^{\infty}\dfrac{1-t^{\frac{1}{n}}}{1-t}t^{\frac{1}{n}-1}\,dt$$
$$I(n)=\dfrac{1}{n}\left[\underbrace{\displaystyle \int_{0}^{\infty}\dfrac{t^{\frac{1}{n}-1}}{1-t}\,dt}_{F_{1}(t)}- \underbrace
{\displaystyle \int_{0}^{\infty}\dfrac{t^{\frac{2}{n}-1}}{1-t}\,dt}_{F_{2}(t)}\right]$$
So again we have two separate integrals , now using melling transform we can evaluate these two very easily , so first lets define the

standard Mellin transform of a function let's say $f(t)$ , it is given by ..
$$\mathcal{M}[f(t)]=F(s)=\displaystyle \int_{0}^{\infty}f(t)t^{s-1}dt$$
So, now if we compare $F_{1}(t)$ with the above integral then we can see that ..
$s_{1}=\dfrac{1}{n}$ and $f_{1}(t)=\dfrac{1}{1-t}$
Now Mellin transform of $f_{1}(t)$ is well known, it is ...
$$\mathcal{M}[f_{1}(t)]\bigg{|}_{s_{1}=1/n}=\pi \cot(\pi s)\bigg{|}_{s=s_{1}}=\pi \cot\left(\dfrac{\pi}{n}\right),0and for $f_{2}(t)$ at $s_{2}=\dfrac{2}{n}$ it will be
$$\mathcal{M}[f_{2}(t)]\bigg{|}_{s_{2}=2/n}=\pi \cot(\pi s)\bigg{|}_{s=s_{2}}=\pi \cot\left(\dfrac{2\pi}{n}\right),0$$I(n)=\dfrac{1}{n} \left[\mathcal{M}[f_{1}(t)]\bigg{|}_{s_{1}=2/n}- \mathcal{M}[f_{2}(t)]\bigg{|}_{s_{2}=2/n} \right]$$
$$I(n)=\dfrac{\pi}{n} \left[\cot\left(\dfrac{\pi}{n}\right)-\cot\left(\dfrac{2\pi}{n}\right)\right]$$

So, again we arrive at the same result i.e
$$\Large \displaystyle\ \bbox[10pt, border:2pt solid #06f]{I(n)=\dfrac{\pi}{n}\csc\left(\dfrac{2 \pi}{n}\right)} \tag*{}$$

elementary set theory - Set of continuous functions from $mathbb{R}$ to $mathbb{R}$ is equinumerous to $mathbb{R}$

I am trying to prove that the set $\mathcal{C}:=\{f\mid f\in\mathbb{R}^\mathbb{R}\mid f \text{ is continuous}\}$ is equinumerous to $\mathbb{R}$.

To achieve this, I note that an $f\in\mathcal C$ is ''pinned down'' by $\mathbb{Q}$, i.e. if I know how $f$ behaves on $\mathbb{Q}$, I know all of $f$, since $\mathbb{Q}$ is dense in $\mathbb{R}$ and $f$ is continuous. So there is an easy injection from $\mathcal{C}$ to $\{f\mid f\in\mathbb{R}^\mathbb{Q}\mid f\text{ is continuous}\}$.

The latter set can be embedded into $\mathbb{R}^\mathbb{N}$, but I'm afraid that fact won't help me, since (as I believe): not $\mathbb{R}^\mathbb{N}\preccurlyeq\mathbb{R}$.

Could someone hint me into the right direction on proving $\mathcal{C}\sim\mathbb{R}$ (via $\mathcal{C}\preccurlyeq\mathbb{R}$)?

Answer

You have $$\vert \mathbb R^{\mathbb N} \vert = \vert (2^{\mathbb N})^{\mathbb N} \vert = \vert 2^{\mathbb N \times \mathbb N} \vert = \vert 2^{\mathbb N} \vert = \vert \mathbb R \vert$$

integration - $int cos^5(x) dx$ with reduction formula

I'm trying to calculate $$\int cos^5(x)dx$$ with the reduction formula.

$$\int cos^5(x)dx=\frac{1}{5}cos^4(x)sin(x)+\frac{4}{5}\int cos^3(x)dx$$
then
$$\int cos^3(x)dx=\frac{1}{3}cos^2(x)sin(x)+\frac{2}{3}\int cos(x)dx$$

so
$$\int cos^5(x)dx=\frac{1}{5}cos^4(x)sin(x)+\frac{4}{5}(\frac{1}{3}cos^2(x)sin(x)+\frac{2}{3}sin(x))$$

But this looks different from:

$$\frac{1}{5}sin^5(x)-\frac{2}{3}sin^3(x)+sin(x)$$

given by:

https://www.symbolab.com/solver/step-by-step/%5Cint%20cos%5E%7B5%7D%5Cleft%28x%5Cright%29dx/?origin=button

Answer

Converting the cosines in your answer to sines via the pythagorean identity, $\cos ^2x=1-\sin ^2x$ yields:

$\frac{1}{5}\cos^4(x)\sin(x)+\frac{4}{5}(\frac{1}{3}\cos^2(x)\sin(x)+\frac{2}{3}\sin(x))$
$=\frac{1}{5}(1-\sin^2(x))^2\sin(x)+\frac{4}{5}(\frac{1}{3}(1-\sin^2(x))\sin(x)+\frac{2}{3}\sin(x))$

Simplifying from there:

$=\frac{1}{5}(1-2\sin^2(x)+\sin^4(x))\sin(x)+\frac{4}{5}(\frac{1}{3}\sin(x)-\frac{1}{3}\sin^3(x))+\frac{2}{3}\sin(x))$

$=\frac{1}{5}\sin(x)-\frac{2}{5}\sin^3(x)+\frac{1}{5}\sin^5(x)+\frac{4}{15}\sin(x)-\frac{4}{15}\sin^3(x))+\frac{8}{15}\sin(x)$

$=\frac{1}{5}\sin^5(x)-\frac{2}{3}\sin^3(x)+\sin(x)$

Which agrees with the answer from the solver.

real analysis - convergence of a recursive sequence with parameter a

How can you determine if the following recursive sequence converges:

$$x_{n+1}=\frac{1}{2}(a+x_n^2)$$

where $0\le a \le 1$ and $x_1=0$

I know that the limit x (if it exists) satisfies the following equation:

$$x=\frac{1}{2}(a+x^2)$$

since $\lim_{x\rightarrow \infty} x_n = \lim_{x\rightarrow \infty} x_{n+1}$. Therefore $x=1\pm \sqrt{1-a}$

Thank you in advance :)

Answer

That is a really useful obervation. Notice that by induction you can easily prove that $x_n\le 1$ which means that the option $1+\sqrt{1-a}$ is definitely not the limit for $a<1$, since then it's strictly greater than 1. If $a=1$ then $1+\sqrt{1-a}=1-\sqrt{1-a}$. That means that if the limit exists, it's equal to $1-\sqrt{1-a}$.

Now, the case for $a=0$ is obvious since then $x_n=0$ for all $n$ so we will assume $a>1$. Notice that if we prove that the sequence is bounded from above and increasing, we will get convergence and from the above argument the limit. We can show inductively that the sequence is bounded from above by $1-\sqrt{1-a}$. The base case $x_0<1-\sqrt{1-a}$ is obviously fulfilled. Now if $x_k<1-\sqrt{1-a}$, then

$x_{k+1}=\frac{1}{2}(a+x_k^2)<\frac{1}{2}(a+(1-\sqrt{1-a})^2)=\frac{1}{2}(a+1-2\sqrt{1-a}+1-a)=1-\sqrt{1-a}$.

That complete the induction. Now to prove the sequence is increasing, you can notice that $x_n<\frac{1}{2}(a+x_n^2)$ for all $0\le x_n<1-\sqrt{1-a}$ (it's a quadratic in $x$) which means $x_{n+1}=\frac{1}{2}(a+x_n^2)>x_n$. That completes the proof that the seqence $(x_n)$ is convergent. As we stated earlier, convergence implies that the limit is $1+\sqrt{1-a}$, which means we're done.

contest math - Inequality, Cauchy Schwarz and Schur

For $a,b, c>0$, prove that $$\frac{a^3}{a^3+b^3+abc}+\frac{b^3}{b^3+c^3+abc}+\frac{c^3}{c^3+a^3+abc}\geq 1$$

I tried the following $$\sum_{cyc}\frac{a^3}{a^3+b^3+abc}\cdot \sum_{cyc}a^3+b^3+abc\geq (a^{3/2}+b^{3/2}+c^{3/2})^2$$
where cyc stands for cyclic sum of $a,b,c$, then we could show that $$(a^{3/2}+b^{3/2}+c^{3/2})^2\geq 2(a^3+b^3+c^3)+3abc$$ to prove our original inequality. This looks similar to Schur but with inequality sign reversed. How would you proceed? Any ideas/hint?

linear algebra - Intuition behind Matrix being invertible iff determinant is non-zero

I have been wondering about this question since I was in school. How can one number tell so much about the whole matrix being invertible or not?

I know the proof of this statement now. But I would like to know the intuition behind this result and why this result is actually true.

---

The proof I have in mind is:

If $A$ is invertible, then

$$ 1 = \det(I) = \det(AA^{-1}) = \det(A)\cdot\det(A^{-1})$$

whence $\det(A) \neq 0$.

Conversely, if $\det(A) \neq 0$, we have

$$ A adj(A) = adj(A)A = \det(A)I$$

whence $A$ is invertible.

$adj(A)$ is the adjugate matrix of $A$.

$$ adj(A)_{ji} = (-1)^{i+j}\det(A_{ij})$$

where $A_{ij}$ is the matrix obtained from $A$ by deleting $ith$ row and $jth$ column.

Any other insightful proofs are also welcome.

sequences and series - Evaluating $limlimits_{ntoinfty} dfrac{1}{n^3}sumlimits_{ell=1}^{n-1}sqrt{(n^2-ell^2)(n^2-(ell-1)^2)}$

Along the way to proving a solution for this stubborn question of mine, I've come upon this expression which I would like to evaluate:
$$
\lim_{n\to\infty} \frac{1}{n^3}\sum_{\ell=1}^{n-1}\sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)}

$$
Assuming consistency+correctness of the rest of my work, I would love for it to turn out that the limit is $\frac{2}{3}$, but to be honest I'm not certain of how to continue. I see no reason for there to be a nice closed form for the sum (and W|A appear to agree).

Answer

For every $1\leqslant\ell\leqslant n-1$,
$$
n^2-\ell^2\leqslant\sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)}\leqslant n^2-(\ell-1)^2,
$$
hence the sums $S_n$ you are interested in are such that $R_n\leqslant S_n\leqslant T_n$ for every $n\geqslant1$, with
$$
R_n=\frac1{n^3}\sum_{\ell=1}^{n-1}(n^2-\ell^2),\qquad T_n=\frac1{n^3}\sum_{\ell=0}^{n-2}(n^2-\ell^2).

$$
The rest should be easy (and the limit is indeed $\frac23$).