Check convergence of integrals

[mathmari]

Hey! :giggle:

I want to check if the following integrals converge or diverge.

1 . $\displaystyle{\int_0^{+\infty}t^2e^{-t^2}\, dt}$
2. $\displaystyle{\int_e^{+\infty}\frac{1}{t^n\ln t}\, dt, \ n\in \{1,2\}}$
3. $\displaystyle{\int_0^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt}$
4. $\displaystyle{\int_0^{+\infty}e^{-t^2}\, dt}$
5. $\displaystyle{\int_1^{+\infty}\frac{(\ln t)^{\beta}}{t^{\alpha}}\, dt, \alpha ,\beta\in \mathbb{R}}$
6. $\displaystyle{\int_0^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt}$
7. $\displaystyle{\int_1^{+\infty}\frac{|\sin t|}{t}\, dt}$
8. $\displaystyle{\int_0^{+\infty}t^ne^{-at}\, dt, \ a>0}$
9. $\displaystyle{\int_1^{+\infty}\frac{|\cos t|}{t^2}\, dt}$ Let's consider first the three first integrals...

For the integral 1 I have done the following:

It holds that $\left |t^2e^{-t^2}\right |\leq te^{-t}$ and that $\displaystyle{\int_0^{+\infty}te^{-t}\, dt=1<\infty}$. Therefore also the integral $\displaystyle{\int_0^{+\infty}t^2e^{-t^2}\, dt}$ converges. For question 2 the integral diverges, or not?So wehave to take a smaller integral that can be computed to show that the smaller one diverges and so also the origonal integral, right? But what bound do we take? For question 3 I thought to take $\left |\frac{\sin t}{\sqrt{t}}\right |=\frac{|\sin t|}{|\sqrt{t}|}\leq \frac{t}{\sqrt{t}}=\sqrt{t}$ but the integral of $\sqrt{t}$ diverges. The same holds if we take $|\sin t|<1$ instead of $|\sin t|<t$. Which bound do we take here? :unsure:

 

[I like Serena]

1 . $\displaystyle{\int_0^{+\infty}t^2e^{-t^2}\, dt}$
For the integral 1 I have done the following:
It holds that $\left |t^2e^{-t^2}\right |\leq te^{-t}$ and that $\displaystyle{\int_0^{+\infty}te^{-t}\, dt=1<\infty}$. Therefore also the integral $\displaystyle{\int_0^{+\infty}t^2e^{-t^2}\, dt}$ converges.

Hey mathmari!

How can we tell that $\left |t^2e^{-t^2}\right |\leq te^{-t}$? :unsure:

2. $\displaystyle{\int_e^{+\infty}\frac{1}{t^n\ln t}\, dt, \ n\in \{1,2\}}$
For question 2 the integral diverges, or not?So wehave to take a smaller integral that can be computed to show that the smaller one diverges and so also the origonal integral, right? But what bound do we take?

How about $\frac{1}{t^\alpha}$? :unsure:

3. $\displaystyle{\int_0^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt}$
For question 3 I thought to take $\left |\frac{\sin t}{\sqrt{t}}\right |=\frac{|\sin t|}{|\sqrt{t}|}\leq \frac{t}{\sqrt{t}}=\sqrt{t}$ but the integral of $\sqrt{t}$ diverges. The same holds if we take $|\sin t|<1$ instead of $|\sin t|<t$. Which bound do we take here?

Suppose we try partial integration first? That is $\int u\,dv = u\,v-\int v\,du$. :unsure:

 

[mathmari]

How can we tell that $\left |t^2e^{-t^2}\right |\leq te^{-t}$? :unsure:

A good question... This proof is a bit difficult because of the powers at the exponential, isn't it? Is there an other upper bound that we could use? :unsure:

How about $\frac{1}{t^\alpha}$? :unsure:

By $a$ you mean $n$ ? :unsure:

Suppose we try partial integration first? That is $\int u\,dv = u\,v-\int v\,du$. :unsure:

But does this help us?

We have \begin{equation*}\int_0^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt=\int_0^{+\infty}\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )'\, dt=\left [\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )\right ]_0^{+\infty}-\int_0^{+\infty}\left (\frac{1}{\sqrt{t}}\right )'\cdot \left (-\cos t\right )\, dt\end{equation*} The limit of $\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )$ at $0$ doesn't exist. :unsure:

 

[I like Serena]

A good question... This proof is a bit difficult because of the powers at the exponential, isn't it? Is there an other upper bound that we could use?

Perhaps we can use $t^2 e^{-t^2} < t^2 e^{-t}$ for sufficiently large $t$? :unsure:

Or alternatively we could do a partial integration first.

By $a$ you mean $n$ ?

I meant a power $\alpha$ that could be anything.
Note that for sufficiently large $t$ we have $\frac 1{t^3} <\frac 1{t^2\ln t} <\frac 1{t^2} <\frac 1{t^{1.01}} < \frac 1{t\ln t} < \frac 1{t}$.

We have \begin{equation*}\int_0^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt=\int_0^{+\infty}\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )'\, dt=\left [\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )\right ]_0^{+\infty}-\int_0^{+\infty}\left (\frac{1}{\sqrt{t}}\right )'\cdot \left (-\cos t\right )\, dt\end{equation*} The limit of $\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )$ at $0$ doesn't exist.

Suppose we split the integral into a part from $0$ to $1$ and a part from $1$ to $\infty$.
Can we find an upper bound for the first part for the original integral?
And an upper bound for the second part after the partial integration?

 

[Opalg]

2. $\displaystyle{\int_e^{+\infty}\frac{1}{t^n\ln t}\, dt, \ n\in \{1,2\}}$

For $n=1$ you can integrate it explicitly: $\displaystyle \int \frac1{t\ln t}\,dt = \ln(\ln t)$.

 

[mathmari]

Perhaps we can use $t^2 e^{-t^2} < t^2 e^{-t}$ for sufficiently large $t$? :unsure:

Or alternatively we could do a partial integration first.

When we do partial integration we get the integral $\int_0^{+\infty}e^{-t^2}\, dt$, right? Do we want to find now for $e^{-t^2}$ an upper bound? :unsure:
But $e^{-t^2}\leq e^{-t}$ holds only for $t\geq 1$. So either we have to find an other upper bound or we have to split the integral into a sum of two integrals, but what could we do then with the integral on $[0,1]$ ?

:unsure:

Suppose we split the integral into a part from $0$ to $1$ and a part from $1$ to $\infty$.
Can we find an upper bound for the first part for the original integral?
And an upper bound for the second part after the partial integration?

We have the following: \begin{equation*}\int_0^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt=\int_0^{1}\frac{\sin t}{\sqrt{t}}\, dt+\int_1^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt\end{equation*}

- $\displaystyle{\int_0^{1}\frac{\sin t}{\sqrt{t}}\, dt}$ :

We have that $\left |\frac{\sin t}{\sqrt{t}}\right |\leq \frac{t}{\sqrt{t}}=\sqrt{t}$ and $\int_0^1 \sqrt{t} \, dt=\left [\frac{2t^{3/2}}{3}\right ]_0^1=\frac{2}{3}<\infty$. So also the integral $\displaystyle{\int_0^{1}\frac{\sin t}{\sqrt{t}}\, dt}$ converges.

- $\displaystyle{\int_1^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt}$ :
Using partial integration we get \begin{align*}\int_1^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt&=\int_1^{+\infty}\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )'\, dt=\left [\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )\right ]_1^{+\infty}-\int_1^{+\infty}\left (\frac{1}{\sqrt{t}}\right )'\cdot \left (-\cos t\right )\, dt \\ & =\lim_{t\rightarrow +\infty}\frac{-\cos t}{\sqrt{t}}-\left (\frac{-\cos 1}{\sqrt{1}}\right )+\int_1^{+\infty}\left (-\frac{1}{2t^{3/2}}\right )\cdot \cos t \, dt \\ & =\cos 1-\int_1^{+\infty} \frac{\cos t}{2t^{3/2}} \, dt\end{align*}
We have that $\left |\frac{\cos t}{2t^{3/2}}\right |\leq \frac{1}{2t^{3/2}}$ und $\int_1^{+\infty} \frac{1}{2t^{3/2}} \, dt=\left [-\frac{1}{\sqrt{t}}\right ]_1^{+\infty}=0+1=1<\infty$. So also the integral $\displaystyle{\int_1^{+\infty} \frac{\cos t}{2t^{3/2}} \, dt}$ converges.

So also the integral $\displaystyle{\int_1^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt}$ converges.

Therefore we get the convergence also of $\displaystyle{\int_0^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt}$.

:unsure:

 

[mathmari]

For $n=1$ you can integrate it explicitly: $\displaystyle \int \frac1{t\ln t}\,dt = \ln(\ln t)$.

So we have to consider the cases $n=1$ and $n=2$ seperately, right? :unsure:

 

[Opalg]

So we have to consider the cases $n=1$ and $n=2$ seperately, right? :unsure:

I think so, yes.

 

[mathmari]

I think so, yes.

Ah ok!

So we have:

For $n=1$ : $\displaystyle \int_e^{+\infty} \frac1{t\ln t}\,dt = \left [\ln(\ln t)\right ]_e^{+\infty}= \lim_{t\rightarrow +\infty}\ln(\ln t)=+\infty$, so we see that this integral doesn't converge.

For $n=2$, using the inequalities of post #4 we get that $\left |\frac{1}{t^2\ln t}\right |\leq \frac{1}{t^2}$ and since $\displaystyle \int_e^{+\infty} \frac{1}{t^2}\,dt =\left [-\frac{1}{t}\right ]_e^{+\infty}=\frac{1}{e}<\infty$. So also the integral $\displaystyle \int_e^{+\infty} \frac1{t^2\ln t}\,dt$ converges.

Is everything correct? :unsure:

 

[Opalg]

Ah ok!

So we have:

For $n=1$ : $\displaystyle \int_e^{+\infty} \frac1{t\ln t}\,dt = \left [\ln(\ln t)\right ]_e^{+\infty}= \lim_{t\rightarrow +\infty}\ln(\ln t)=+\infty$, so we see that this integral doesn't converge.

For $n=2$, using the inequalities of post #4 we get that $\left |\frac{1}{t^2\ln t}\right |\leq \frac{1}{t^2}$ and since $\displaystyle \int_e^{+\infty} \frac{1}{t^2}\,dt =\left [-\frac{1}{t}\right ]_e^{+\infty}=\frac{1}{e}<\infty$. So also the integral $\displaystyle \int_e^{+\infty} \frac1{t^2\ln t}\,dt$ converges.

Is everything correct? :unsure:

(up)

 

[I like Serena]

When we do partial integration we get the integral $\int_0^{+\infty}e^{-t^2}\, dt$, right? Do we want to find now for $e^{-t^2}$ an upper bound? :unsure:
But $e^{-t^2}\leq e^{-t}$ holds only for $t\geq 1$. So either we have to find an other upper bound or we have to split the integral into a sum of two integrals, but what could we do then with the integral on $[0,1]$ ?

The integral of a bounded function on a bounded interval is finite isn't it?

We have the following: \begin{equation*}\int_0^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt=\int_0^{1}\frac{\sin t}{\sqrt{t}}\, dt+\int_1^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt\end{equation*}

- $\displaystyle{\int_0^{1}\frac{\sin t}{\sqrt{t}}\, dt}$ :

We have that $\left |\frac{\sin t}{\sqrt{t}}\right |\leq \frac{t}{\sqrt{t}}=\sqrt{t}$ and $\int_0^1 \sqrt{t} \, dt=\left [\frac{2t^{3/2}}{3}\right ]_0^1=\frac{2}{3}<\infty$. So also the integral $\displaystyle{\int_0^{1}\frac{\sin t}{\sqrt{t}}\, dt}$ converges.

- $\displaystyle{\int_1^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt}$ :
Using partial integration we get \begin{align*}\int_1^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt&=\int_1^{+\infty}\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )'\, dt=\left [\frac{1}{\sqrt{t}}\cdot \left (-\cos t\right )\right ]_1^{+\infty}-\int_1^{+\infty}\left (\frac{1}{\sqrt{t}}\right )'\cdot \left (-\cos t\right )\, dt \\ & =\lim_{t\rightarrow +\infty}\frac{-\cos t}{\sqrt{t}}-\left (\frac{-\cos 1}{\sqrt{1}}\right )+\int_1^{+\infty}\left (-\frac{1}{2t^{3/2}}\right )\cdot \cos t \, dt \\ & =\cos 1-\int_1^{+\infty} \frac{\cos t}{2t^{3/2}} \, dt\end{align*}
We have that $\left |\frac{\cos t}{2t^{3/2}}\right |\leq \frac{1}{2t^{3/2}}$ und $\int_1^{+\infty} \frac{1}{2t^{3/2}} \, dt=\left [-\frac{1}{\sqrt{t}}\right ]_1^{+\infty}=0+1=1<\infty$. So also the integral $\displaystyle{\int_1^{+\infty} \frac{\cos t}{2t^{3/2}} \, dt}$ converges.

So also the integral $\displaystyle{\int_1^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt}$ converges.

Therefore we get the convergence also of $\displaystyle{\int_0^{+\infty}\frac{\sin t}{\sqrt{t}}\, dt}$.

Looks good to me. (Nod)

 

[mathmari]

The integral of a bounded function on a bounded interval is finite isn't it?

Using partial integration we get \begin{align*}\int_0^{+\infty}t^2e^{-t^2}\, dt&=\int_0^{+\infty}t\cdot \frac{1}{-2}\cdot \left (-2te^{-t^2}\right )\, dt =-\frac{1}{2}\cdot \int_0^{+\infty}t\cdot \left (e^{-t^2}\right )'\, dt\\ & =-\frac{1}{2}\cdot \left (\left [t\cdot e^{-t^2}\right ]_0^{+\infty}- \int_0^{+\infty}\left (t\right )'\cdot e^{-t^2}\, dt\right ) \\ & =-\frac{1}{2}\cdot \left (\lim_{t\rightarrow +\infty}t\cdot e^{-t^2}-0\cdot e^{-0^2}- \int_0^{+\infty}1\cdot e^{-t^2}\, dt\right ) \\ & =-\frac{1}{2}\cdot \left (\lim_{t\rightarrow +\infty}\frac{t}{e^{t^2}}- \int_0^{+\infty} e^{-t^2}\, dt\right ) \\ & \ \overset{\text{De L'Hospital}}{ = } \ -\frac{1}{2}\cdot \left (\lim_{t\rightarrow +\infty}\frac{1}{2te^{t^2}}- \int_0^{+\infty} e^{-t^2}\, dt\right ) \\ & = -\frac{1}{2}\cdot \left (0- \int_0^{+\infty} e^{-t^2}\, dt\right ) \\ & = \frac{1}{2}\cdot\int_0^{+\infty} e^{-t^2}\, dt \\ & = \frac{1}{2}\cdot \left (\int_0^1 e^{-t^2}\, dt+\int_1^{+\infty} e^{-t^2}\, dt\right )\end{align*}
It holds that \begin{equation*}0<t<1 \Rightarrow 0<t^2<1 \Rightarrow -1<-t^2<0 \Rightarrow e^{-1}<e^{-t^2}<e^0 \Rightarrow \frac{1}{e}<e^{-t^2}<1\end{equation*}
The integral of a bounded function on a bounded interval is finite. So we have that $\displaystyle{\int_0^1 e^{-t^2}\, dt<\infty}$.

For $t\geq 1$ it holds that $\displaystyle{|e^{-t^2}|=e^{-t^2}\leq e^{-t}}$ and $\displaystyle{\int_1^{+\infty} e^{-t}\, dt=\left [ -e^{-t}\right ]_1^{+\infty}=\lim_{t\rightarrow +\infty}\left ( -e^{-t}\right )- \left ( -e^{-1}\right )=-\lim_{t\rightarrow +\infty}\frac{1}{e^{t}}+\frac{1}{e}=-0+\frac{1}{e}=\frac{1}{e}<\infty}$.
So also the integral $\displaystyle{\int_1^{+\infty} e^{-t^2}\, dt}$ converges.

Therefore we get the convergence of $\displaystyle{\int_0^{+\infty}t^2e^{-t^2}\, dt}$. Is that correct? :unsure:

 

[mathmari]

At integral #5 do we have to take cases for $\alpha$ and $\beta$ ?At integral #6 :
Do we use that $$\left |\frac{t\ln t}{(1-t^2)^2}\right | \leq \frac{t \cdot t}{(1-t^2)^2}=\frac{t^2}{(1-t^2)^2}$$ Or can we get a better upper bound?
At integral #7 :
We have that $$\left |\frac{|\sin t|}{t}\right |=\frac{|\sin t|}{t}\leq \frac{t}{t}=1$$ But that doesn't help us, since the integral of $1$ doesn't converge.At integral #8 :
Do we use partial integration?At integral #9 :
We have that $\displaystyle{\left |\frac{|\cos t|}{t^2}\right |=\frac{|\cos t|}{t^2}\leq \frac{1}{t^2}}$ and $\displaystyle{\int_1^{+\infty}\frac{1}{t^2}\, dt=\left [-\frac{1}{t}\right ]_1^{+\infty}=1<\infty}$. So we get also convergence of $\displaystyle{\int_1^{+\infty}\frac{|\cos t|}{t^2}\, dt}$.:unsure:

 

[Theia]

Nothing new, but two other thoughts about integral 3:

After substituting \( t=x^2 \) one obtains the integral \[ \textrm{constant} \cdot \int_0^{\infty} \sin (x^2) \textrm{d}x, \] which is a standard integral known to converge.Another way to proceed:

\[ \begin{align*}\int_0^{\infty} \frac{\sin (t)}{\sqrt{t}}\textrm{d}t &= \sum_{v=0}^{\infty} \int_{v\pi}^{(v+1)\pi} \frac{\sin (t)}{\sqrt{t}} \textrm{d}t \\
&< \pi + \sum_{v=1}^{\infty} \frac{(-1)^v}{\sqrt{v\pi}} \\
&< \pi + \frac{1}{\sqrt{2\pi}},\end{align*}\]

because alternating sum converges.

 

[mathmari]

At integral #6 I have done the following:

We have that \begin{equation*}\int_0^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt=\int_0^1\frac{t\ln t}{(1-t^2)^2}\, dt+\int_1^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt\end{equation*}

In the interval $[0,1]$ we have that \begin{equation*}\left |\frac{t\ln t}{(1+t^2)^2}\right |\leq \frac{t\cdot \frac{1}{t}}{t^4}=\frac{1}{t^4}<1\end{equation*}
The integral of a bounded function in a bounded intervall is finite.

In the Interval $[1, +\infty)$ we have that \begin{equation*}\left |\frac{t\ln t}{(1+t^2)^2}\right |\leq \frac{t\cdot t}{t^4}=\frac{t^{2}}{t^4}=t^{-2}\end{equation*} and \begin{equation*}\int_1^{+\infty}t^{-2}\, dt=\left [-\frac{1}{t}\right ]_1^{+\infty}=\lim_{t\rightarrow +\infty}\left (-\frac{1}{t}\right )-\left (-\frac{1}{1}\right )=0+1=1\end{equation*}
So also the integral $\displaystyle{\int_1^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt}$ converges.

Therefore we get also the convergence of $\displaystyle{\int_0^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt}$.Is that correct? :unsure:

 

[mathmari]

At integral #7 I have done the following:

We have that \begin{equation*}\displaystyle \int_1^\infty \frac{|\sin(x)|}{x} \,\mathrm{d}x = \int_1^\pi \frac{|\sin(x)|}{x} \,\mathrm{d}x + \sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{x} \,\mathrm{d}\end{equation*}
The first term is finite, so we have to check the convergnce of the series.

It holds that \begin{align*}\sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{x} \,\mathrm{d}x &\geq \sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{(k+1)\pi} \,\mathrm{d}x = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\int_{k\pi}^{(k+1)\pi} |\sin(x)| \,\mathrm{d}x \\ & = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\left [-\text{sgn}(\sin(x))\cdot \cos (x)\right ]_{k\pi}^{(k+1)\pi}\end{align*}
Is that correct? Do we get now the harmonic series?

:unsure:

 

[I like Serena]

Is that correct?

Looks correct to me. (Nod)

5. $\displaystyle{\int_1^{+\infty}\frac{(\ln t)^{\beta}}{t^{\alpha}}\, dt, \alpha ,\beta\in \mathbb{R}}$
At integral #5 do we have to take cases for $\alpha$ and $\beta$ ?

For any $\alpha,\beta >0$ we have that $(\ln t)^\beta < t^\alpha$ for sufficiently large $t$.
Additionally we know that $\int_1^{+\infty}\frac 1 t\,dt$ diverges and that for $\alpha>0$ that $\int_1^{+\infty}\frac 1 {t^{1+\alpha}}\,dt$ converges.
So I guess we have to find where the boundaries are for $\alpha$ and $\beta$ so that we get either convergence or divergence.

6. $\displaystyle{\int_0^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt}$

We have $$\frac{t^2}{(1-t^2)^2} = \frac{1}{(1-t^2)^2} - \frac{1-t^2}{(1-t^2)^2}$$
don't we?

Btw, we can't just replace $(1-t^2)^2$ by $(1+t^2)^2$ can we?

7. $\displaystyle{\int_1^{+\infty}\frac{|\sin t|}{t}\, dt}$
At integral #7 I have done the following:

We have that \begin{equation*}\displaystyle \int_1^\infty \frac{|\sin(x)|}{x} \,\mathrm{d}x = \int_1^\pi \frac{|\sin(x)|}{x} \,\mathrm{d}x + \sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{x} \,\mathrm{d}\end{equation*}
The first term is finite, so we have to check the convergnce of the series.

It holds that \begin{align*}\sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{x} \,\mathrm{d}x &\geq \sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{(k+1)\pi} \,\mathrm{d}x = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\int_{k\pi}^{(k+1)\pi} |\sin(x)| \,\mathrm{d}x \\ & = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\left [-\text{sgn}(\sin(x))\cdot \cos (x)\right ]_{k\pi}^{(k+1)\pi}\end{align*}
Is that correct? Do we get now the harmonic series?

Yes, we get the harmonic series since the integrated part can be made independent of $k$.

8. $\displaystyle{\int_0^{+\infty}t^ne^{-at}\, dt, \ a>0}$
At integral #8 :
Do we use partial integration?

Let's try. (Sweating)

9. $\displaystyle{\int_1^{+\infty}\frac{|\cos t|}{t^2}\, dt}$
At integral #9 :
We have that $\displaystyle{\left |\frac{|\cos t|}{t^2}\right |=\frac{|\cos t|}{t^2}\leq \frac{1}{t^2}}$ and $\displaystyle{\int_1^{+\infty}\frac{1}{t^2}\, dt=\left [-\frac{1}{t}\right ]_1^{+\infty}=1<\infty}$. So we get also convergence of $\displaystyle{\int_1^{+\infty}\frac{|\cos t|}{t^2}\, dt}$.

Yep. (Nod)

 

[mathmari]

For any $\alpha,\beta >0$ we have that $(\ln t)^\beta < t^\alpha$ for sufficiently large $t$.
Additionally we know that $\int_1^{+\infty}\frac 1 t\,dt$ diverges and that for $\alpha>0$ that $\int_1^{+\infty}\frac 1 {t^{1+\alpha}}\,dt$ converges.
So I guess we have to find where the boundaries are for $\alpha$ and $\beta$ so that we get either convergence or divergence.

For $\alpha,\beta >0$ do we have that $(\ln t)^\beta < t^\alpha$ if $\beta \leq \alpha$ ? :unsure:
We have that $(\ln t)^\beta < t^\alpha< t^{\alpha+2}$ and so we get $\frac{(\ln t)^\beta}{t^\alpha}<\frac{1}{t^2}$ and since $\int_1^{+\infty}\frac 1 {t^2}\,dt=\left [-\frac{1}{t}\right ]_1^{+\infty}=1<\infty$ and so we get convergence also for the integral $\displaystyle{\int_1^{+\infty}\frac{(\ln t)^{\beta}}{t^{\alpha}}\, dt}$.

So now we have to check also the cases "$\alpha>0$ & $\beta<0$", "$\alpha<0$ & $\beta>0$" and "$\alpha<0$ & $\beta<0$", right? :unsure:

We have $$\frac{t^2}{(1-t^2)^2} = \frac{1}{(1-t^2)^2} - \frac{1-t^2}{(1-t^2)^2}$$
don't we?

Btw, we can't just replace $(1-t^2)^2$ by $(1+t^2)^2$ can we?

Ah yes... I thought my mistake that we had "+" instead of "-". (Tmi)

So in the interval $[1, +\infty)$ we have $$\left |\frac{t\ln t}{(1-t^2)^2}\right |\leq \frac{t\cdot t}{(1-t^2)^2}=\frac{t^2}{(1-t^2)^2}= \frac{1}{(1-t^2)^2} - \frac{1-t^2}{(1-t^2)^2}= \frac{1}{(1-t^2)^2} - \frac{1}{1-t^2}$$ The second part is positive, isn't it? :unsure:

Yes, we get the harmonic series since the integrated part can be made independent of $k$.

We have that \begin{align*}\left [-\text{sgn}(\sin(x))\cdot \cos (x)\right ]_{k\pi}^{(k+1)\pi}&=\left (-\text{sgn}(\sin((k+1)\pi))\cdot \cos ((k+1)\pi)\right )-\left (-\text{sgn}(\sin(k\pi))\cdot \cos (k\pi)\right )\\ & =-\text{sgn}(\sin((k+1)\pi))\cdot (-1)^{k+1}+\text{sgn}(\sin(k\pi))\cdot (-1)^k\end{align*}
Since $\sin((k+1)\pi)=\sin(k\pi)=0$ we get that $\text{sgn}(\sin((k+1)\pi))=\text{sgn}(\sin(k\pi))=0$, or not? So do we get the series of $0$ ? :unsure:

Let's try. (Sweating)

I think that first we have to split the integral into the sum $$\int_0^{+\infty}t^ne^{-at}\, dt=\int_0^1t^ne^{-at}\, dt+\int_1^{+\infty}t^ne^{-at}\, dt$$

We have that $\left |\frac{t^n}{e^{at}}\right |=\frac{t^n}{e^{at}}\leq \frac{1^n}{e^{a\cdot 0}}=1$, since $t\in [0,1]$, or not?

Is that is correct, then we have that the integral of a bounded function on a bounded interval is finite.

:unsure: We have that \begin{align*}\int_1^{+\infty}t^ne^{-at}\, dt&=\int_1^{+\infty} \frac{t^n}{-a} \left (e^{-at}\right )'\, dt \\ & =\left [\frac{t^n}{-a} e^{-at}\right ]_1^{+\infty}-\int_1^{+\infty} \left (\frac{t^n}{-a}\right )' e^{-at}\, dt\\ & =\frac{1}{ae^a} -\int_1^{+\infty} \frac{nt^{n-1}}{-a} e^{-at}\, dt \\ & =\frac{1}{ae^a} -\int_1^{+\infty} \frac{nt^{n-1}}{a^2} \left (e^{-at}\right )'\, dt \\ & =\frac{1}{ae^a} -\frac{n}{ae^a}+\int_1^{+\infty} \frac{n(n-1)t^{n-2}}{a^2}e^{-at}\, dt\end{align*}
We if continue we get at the end a sum of the form $\frac{1}{ae^a} -\frac{n}{ae^a}+\frac{n(n-1)}{ae^a} -\frac{n(n-1)(n_2)}{ae^a}+\cdots+\int_1^{+\infty} \frac{n!}{a^n}e^{-at}\, dt$ which can be calculated, or not?

Or is it better to get an upper bound to show the convergence?

:unsure:

 

[I like Serena]

So in the interval $[1, +\infty)$ we have $$\left |\frac{t\ln t}{(1-t^2)^2}\right |\leq \frac{t\cdot t}{(1-t^2)^2}=\frac{t^2}{(1-t^2)^2}= \frac{1}{(1-t^2)^2} - \frac{1-t^2}{(1-t^2)^2}= \frac{1}{(1-t^2)^2} - \frac{1}{1-t^2}$$ The second part is positive, isn't it?

It must be positive yes.
But more importantly we now have 2 terms that we can integrate separately.
If both integrations are finite, then their difference is finite as well.

We have that \begin{align*}\left [-\text{sgn}(\sin(x))\cdot \cos (x)\right ]_{k\pi}^{(k+1)\pi}&=\left (-\text{sgn}(\sin((k+1)\pi))\cdot \cos ((k+1)\pi)\right )-\left (-\text{sgn}(\sin(k\pi))\cdot \cos (k\pi)\right )\\ & =-\text{sgn}(\sin((k+1)\pi))\cdot (-1)^{k+1}+\text{sgn}(\sin(k\pi))\cdot (-1)^k\end{align*}
Since $\sin((k+1)\pi)=\sin(k\pi)=0$ we get that $\text{sgn}(\sin((k+1)\pi))=\text{sgn}(\sin(k\pi))=0$, or not? So do we get the series of $0$ ?

The sign function is a bit confusing here. It is not actually supposed to result in an evaluation of $0$ at the end points.
Perhaps it's better to avoid the sign function and instead integrate separately on $[2\pi k, 2\pi k + \pi]$ and $[2\pi k + \pi, 2\pi k + 2\pi]$.
Then we don't need the sign function and we'll see that the contributions are indeed not zero.

We if continue we get at the end a sum of the form $\frac{1}{ae^a} -\frac{n}{ae^a}+\frac{n(n-1)}{ae^a} -\frac{n(n-1)(n_2)}{ae^a}+\cdots+\int_1^{+\infty} \frac{n!}{a^n}e^{-at}\, dt$ which can be calculated, or not?

Or is it better to get an upper bound to show the convergence?

Suppose we don't split the domain but only apply partial integration.
Don't we get a simpler series then?
We can also evaluate the integral at the end, can't we?

 

[mathmari]

It must be positive yes.
But more importantly we now have 2 terms that we can integrate separately.
If both integrations are finite, then their difference is finite as well.

I checked the integral in Wolfram but this doesn't converge. So we have to find an other upper bound, right? :unsure:

 

[I like Serena]

I checked the integral in Wolfram but this doesn't converge. So we have to find an other upper bound, right?

Ah, I overlooked that we have a singularity at $t=1$.
If we integrate starting from $t=2$ we do have convergence.
I think that instead of an upper bound we need to find a lower bound to prove that e.g. the integral from $1$ to $2$ does not converge.

 

[mathmari]

Ah, I overlooked that we have a singularity at $t=1$.
If we integrate starting from $t=2$ we do have convergence.
I think that instead of an upper bound we need to find a lower bound to prove that e.g. the integral from $1$ to $2$ does not converge.

I don't really find an appropriate lower bound that helps us.

For example in the intervall $[0,1]$ a lower bound is $t\ln t$.
Then it holds also $\ln t\leq t$ for $t\geq 1$ or $\ln t\leq \frac{1}{t}$ for $t\in [0,1]$.

These inequalities are correct, aren't they? :unsure:

But using these ones we don't get an appropriate result, do we?

Could you give me a hint?

:unsure:

 

[I like Serena]

I don't really find an appropriate lower bound that helps us.

For example in the intervall $[0,1]$ a lower bound is $t\ln t$.
Then it holds also $\ln t\leq t$ for $t\geq 1$ or $\ln t\leq \frac{1}{t}$ for $t\in [0,1]$.

$\ln t$ is negative In the interval $(0,1)$ and the function is negative as well, which makes it a bit confusing.

Suppose we focus on the interval $[1,2]$.
Then $\ln t \ge \frac 12(t-1)$ isn't it?

 

[Theia]

If \( t \approx 1 \), then \[ \ln t > 2 \frac{t-1}{t+1}. \]

Which means

\[ \frac{t \ln t}{(1 - t^2)^2} > \frac{2t}{(1+t)^2(1-t)^2} \frac{t-1}{t+1} = \frac{-2t}{(t+1)^3(1-t)} . \]

Does this help? :unsure:

 

[mathmari]

If \( t \approx 1 \), then \[ \ln t > 2 \frac{t-1}{t+1}. \]

Why does this inequality hold? :unsure:

Which means

\[ \frac{t \ln t}{(1 - t^2)^2} > \frac{2t}{(1+t)^2(1-t)^2} \frac{t-1}{t+1} = \frac{-2t}{(t+1)^3(1-t)} . \]

Does this help? :unsure:

So we use this inequality in the interval $[0,1]$ or $[1,2]$ and we get divergence, right? :unsure:

 

[mathmari]

The sign function is a bit confusing here. It is not actually supposed to result in an evaluation of $0$ at the end points.
Perhaps it's better to avoid the sign function and instead integrate separately on $[2\pi k, 2\pi k + \pi]$ and $[2\pi k + k, 2\pi k + 2\pi]$.
Then we don't need the sign function and we'll see that the contributions are indeed not zero.

We have that \begin{align*}\sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{x} \,\mathrm{d}x &\geq \sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{(k+1)\pi} \,\mathrm{d}x = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\int_{k\pi}^{(k+1)\pi} |\sin(x)| \,\mathrm{d}x \\ & = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\left (\int_{(2k-1)\pi}^{(2k)\pi} |\sin(x)| \,\mathrm{d}x +\int_{(2k)\pi}^{(2k+1)\pi} |\sin(x)| \,\mathrm{d}x \right ) \\ & = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\left (\int_{(2k-1)\pi}^{(2k)\pi} \left (-\sin(x)\right ) \,\mathrm{d}x +\int_{(2k)\pi}^{(2k+1)\pi} \sin(x) \,\mathrm{d}x \right ) \\ & = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\left ( \left [\cos(x)\right ]_{(2k-1)\pi}^{(2k)\pi} + \left [-\cos(x)\right ]_{(2k)\pi}^{(2k+1)\pi} \right ) \\ & = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\left ( \left [\cos(x)\right ]_{(2k-1)\pi}^{(2k)\pi} - \left [\cos(x)\right ]_{(2k)\pi}^{(2k+1)\pi} \right ) \\ & = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\left (\left [ (-1)^{2k}-(-1)^{(2k-1)} \right ] -\left [ (-1)^{2k+1}-(-1)^{(2k)} \right ] \right ) \\ & = \frac{1}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\cdot 4 \\ & = \frac{4}{\pi} \sum_{k=1}^\infty \frac{1}{k+1}\end{align*}
The harmonic series is $\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k}}$ we have the series that starts from the second term, right?
So we have that $$k<k+1 \Rightarrow \frac{1}{k}>\frac{1}{k+1} \Rightarrow \sum_{k=1}^\infty \frac{1}{k}> \sum_{k=1}^\infty \frac{1}{k+1}$$ Since $\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k}}$ diverges then it doesn't follow that the smaller one $\displaystyle{\sum_{k=1}^{\infty}\frac{1}{k+1}}$ diverges, does it?

:unsure:

 

[mathmari]

Suppose we don't split the domain but only apply partial integration.
Don't we get a simpler series then?
We can also evaluate the integral at the end, can't we?

We have the following:
\begin{align*}\int_0^{+\infty}t^ne^{-at}\, dt&=\int_0^{+\infty} \frac{t^n}{-a} \left (e^{-at}\right )'\, dt \\ & =\left [\frac{t^n}{-a} e^{-at}\right ]_0^{+\infty}-\int_1^{+\infty} \left (\frac{t^n}{-a}\right )' e^{-at}\, dt\\ & =0 -\int_0^{+\infty} \frac{nt^{n-1}}{-a} e^{-at}\, dt \\ & = -\int_0^{+\infty} \frac{nt^{n-1}}{a^2} \left (e^{-at}\right )'\, dt \\ & =\int_0^{+\infty} \frac{n(n-1)t^{n-2}}{a^2}e^{-at}\, dt \\ & = \ldots \\ & = \int_0^{+\infty} \frac{n!}{a^n}e^{-at}\, dt \\ & = \frac{n!}{a^n}\cdot \int_0^{+\infty} e^{-at}\, dt \\ & = \frac{n!}{a^n}\cdot \left [e^{-at}\right ]_0^{+\infty} \\ & = -\frac{n!}{a^n}<\infty\end{align*}
So the integral converges.

Is that correct? :unsure:

 

[Theia]

Why does this inequality hold? :unsure:

A small correction to my post: must be \( t>1. \) For \( t < 1 \) the > sign turns to <.
The reason why this holds, is the series representation of logarithm:
\[ \ln x = 2\sum_{v=0}^{\infty} \frac{1}{2v+1} \left( \frac{x-1}{x+1} \right)^{2v+1} \]

So we use this inequality in the interval $[0,1]$ or $[1,2]$ and we get divergence, right? :unsure:

Yes.

 

[mathmari]

A small correction to my post: must be \( t>1. \) For \( t < 1 \) the > sign turns to <.
The reason why this holds, is the series representation of logarithm:
\[ \ln x = 2\sum_{v=0}^{\infty} \frac{1}{2v+1} \left( \frac{x-1}{x+1} \right)^{2v+1} \]

Yes.

Sowe have that \begin{equation*}\int_0^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt=\int_0^1\frac{t\ln t}{(1-t^2)^2}\, dt+\int_1^2\frac{t\ln t}{(1-t^2)^2}\, dt+\int_2^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt\end{equation*}
In the interval $[1, 2]$ we have that \begin{equation*}\ln t = 2\sum_{v=0}^{\infty} \frac{1}{2v+1} \left( \frac{t-1}{t+1} \right)^{2v+1}\geq 2\cdot \frac{t-1}{t+1} \end{equation*}
That's why we have that:
\begin{align*}\frac{t\ln t}{(1-t^2)^2}&=\frac{t}{(1-t^2)^2}\cdot \ln t\geq \frac{t}{(1-t^2)^2}\cdot 2\cdot \frac{t-1}{t+1}=\frac{2t}{\left ((1-t)(1+t)\right )^2}\cdot \frac{t-1}{t+1} \\ & =\frac{2t}{(1-t)^2(1+t)^2} \cdot \left (-\frac{1-t}{t+1}\right ) =-\frac{2t}{(1-t)(1+t)^3} \end{align*} und \begin{align*}\int_1^2\frac{-2t}{(1-t)(1+t)^3}\, dt=& \int_1^2-\left (-\frac{1}{4 (-1 + t)} - \frac{1}{(1 + t)^3} + \frac{1}{2 (1 + t)^2} + \frac{1}{4 (1 + t)}\right ) \, dt \\ =& \int_1^2\frac{1}{4 (-1 + t)}\, dt + \int_1^2\frac{1}{(1 + t)^3}\, dt - \int_1^2\frac{1}{2 (1 + t)^2}\, dt - \int_1^2\frac{1}{4 (1 + t)}] \, dt \\ = & \left [\frac{1}{4}\ln (t-1)\right ]_1^2 + \left [-\frac{1}{2(t+1)^2}\right ]_1^2 - \left [-\frac{1}{2(t+1)}\right ]_1^2 - \left [\frac{1}{4}\ln (t+1)\right ]_1^2 \\ = & \frac{1}{4}\ln (2-1)-\lim_{t\rightarrow 1}\frac{1}{4}\ln (t-1) - \frac{1}{2(2+1)^2}+\frac{1}{2(1+1)^2} \\ & +\frac{1}{2(2+1)}-\frac{1}{2(1+1)} - \frac{1}{4}\ln (2+1)+\lim_{t\rightarrow 1}\frac{1}{4}\ln (t+1) \\ = & \frac{1}{4}\ln (3)-\lim_{t\rightarrow 1}\frac{1}{4}\ln (t-1) - \frac{1}{18}+\frac{1}{8} +\frac{1}{6}-\frac{1}{4} - \frac{1}{4}\ln (3)+\lim_{t\rightarrow 1}\frac{1}{4}\ln (t+1) \\ = & -\lim_{t\rightarrow 1}\frac{1}{4}\ln (t-1) +\lim_{t\rightarrow 1}\frac{1}{4}\ln (t+1)-\frac{1}{72}
\\ = & \frac{1}{4}\lim_{t\rightarrow 1}\left (\ln (t+1)-\ln (t-1)\right )-\frac{1}{72} \\ = & \frac{1}{4}\cdot \lim_{t\rightarrow 1}\ln \frac{t+1}{t-1}-\frac{1}{72} \\ = & \frac{1}{4}\cdot \infty -\frac{1}{72} \\ = & \infty \end{align*}
So also the integral $\displaystyle{\int_1^2\frac{t\ln t}{(1-t^2)^2}\, dt}$ doesn't converge.

Therefore neither $\displaystyle{\int_0^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt}$ converges.

Is that correct? :unsure:

 

[mathmari]

As for integral #5 :

We apply the substituion $x=e^t$ ($ \Rightarrow \ln x=t$) and get \begin{equation*}\int_1^{+\infty}\frac{\left (\ln x\right )^{\beta}}{x^{\alpha}}\, dx=\int_0^{+\infty}\frac{\left (t\right )^{\beta}}{\left (e^t\right )^{\alpha}}\, e^t\, dt=\int_0^{+\infty}t^{\beta}e^{(1-\alpha)t}\, dt\end{equation*}

For $1-\alpha\geq 0 \Rightarrow \alpha \leq 1$ we have that $e^{(1-\alpha)t}\geq 1$, so we get:
\begin{equation*}\int_0^\infty t^\beta e^{(1-\alpha)t}dt\geq \int_0^\infty t^\beta dt\rightarrow +\infty\end{equation*}
Therefore also $\displaystyle{\int_1^{+\infty}\frac{\left (\ln x\right )^{\beta}}{x^{\alpha}}\, dx}$ diverges for $\beta\in \mathbb{R}$.

For $1-\alpha <0 \Rightarrow \alpha > 1$ we have that:
\begin{equation*}\int_0^{+\infty}t^{\beta}e^{(1-\alpha)t}\, dt=\int_0^1t^{\beta}e^{(1-\alpha)t}\, dt+\int_1^{+\infty}t^{\beta}e^{(1-\alpha)t}\, dt\end{equation*}

- $\displaystyle{\int_0^1t^{\beta}e^{(1-\alpha)t}\, dt\leq \int_0^1t^{\beta}\cdot 1\, dt=\int_0^1t^{\beta}\, dt=\left [\frac{t^{\beta+1}}{\beta+1}\right ]_0^1}$

If $\beta+1> 0 \Rightarrow \beta>-1$ then this is equal to $\displaystyle{\frac{1^{\beta+1}}{\beta+1}-\frac{0^{\beta+1}}{\beta+1}=\frac{1}{\beta+1}\in \mathbb{R}}$, so the intagral $\displaystyle{\int_0^1t^{\beta}\, dt}$ converges andd so also $\displaystyle{\int_0^1t^{\beta}e^{(1-\alpha)t}\, dt}$. - We have that $\displaystyle{\int_1^{+\infty}t^{\beta}e^{(1-\alpha)t}\, dt< \int_1^{+\infty}t^{\beta}\, dt=\left [\frac{t^{\beta+1}}{\beta+1}\right ]_1^{+\infty}}$.

If $\beta+1< 0 \Rightarrow \beta<-1$ then this is equal to $\displaystyle{\lim_{t\rightarrow +\infty}\frac{t^{\beta+1}}{\beta+1}-\frac{1^{\beta+1}}{\beta+1}=0-\frac{1}{\beta+1}=-\frac{1}{\beta+1}\in \mathbb{R}}$, so the integral $\displaystyle{\int_1^{+\infty}t^{\beta}\, dt}$ converges and so also also $\displaystyle{\int_1^{+\infty}t^{\beta}e^{(1-\alpha)t}\, dt}$. So we have to check the convergence of the second integral if $\beta>-1$ and the convergence of the first integral if $\beta<-1$, or not? :unsure: