Do These Integrals Converge or Diverge?

In summary, the integrals 1, 2, and 3 are all convergent. For the first integral, the upper bound of $te^{-t}$ was used to show convergence. For the second integral, the upper bound of $\frac{1}{t^{\alpha}}$ was used, with $\alpha$ being any power. And for the third integral, the method of partial integration was used to split the integral and find an upper bound for each part, leading to the overall convergence of the integral.
  • #1
mathmari
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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:
 
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  • #2
mathmari said:
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:

mathmari said:
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:

mathmari said:
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:
 
  • #3
Klaas van Aarsen said:
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:
Klaas van Aarsen said:
How about $\frac{1}{t^\alpha}$? :unsure:

By $a$ you mean $n$ ? :unsure:
Klaas van Aarsen said:
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:
 
  • #4
mathmari said:
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. 🤔
mathmari said:
By $a$ you mean $n$ ?

I meant a power $\alpha$ that could be anything.
Note that for sufficiently large $t$ we have $\frac 1{t^3} <\frac 1{t^2\ln t} <\frac 1{t^2} <\frac 1{t^{1.01}} < \frac 1{t\ln t} < \frac 1{t}$. 🤔
mathmari said:
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? 🤔
 
  • #5
mathmari said:
2. $\displaystyle{\int_e^{+\infty}\frac{1}{t^n\ln t}\, dt, \ n\in \{1,2\}}$
For $n=1$ you can integrate it explicitly: $\displaystyle \int \frac1{t\ln t}\,dt = \ln(\ln t)$.
 
  • #6
Klaas van Aarsen said:
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:
Klaas van Aarsen said:
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:
 
  • #7
Opalg said:
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:
 
  • #8
mathmari said:
So we have to consider the cases $n=1$ and $n=2$ seperately, right? :unsure:
I think so, yes.
 
  • #9
Opalg said:
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:
 
  • #10
mathmari said:
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)
 
  • #11
mathmari said:
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? 🤔

mathmari said:
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)
 
  • #12
Klaas van Aarsen said:
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:
 
  • #13
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:
 
  • #14
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.
 
  • #15
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:
 
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  • #16
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:
 
  • #17
mathmari said:
Is that correct?
Looks correct to me. (Nod)

mathmari said:
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. 🤔
mathmari said:
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? :oops:

mathmari said:
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 $k$. 🤔

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

Let's try. (Sweating)

mathmari said:
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)
 
  • #18
Klaas van Aarsen said:
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:
Klaas van Aarsen said:
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? :oops:

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:
Klaas van Aarsen said:
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:

Klaas van Aarsen said:
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:
 
  • #19
mathmari said:
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. 🤔
mathmari said:
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. 🤔
mathmari said:
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? 🤔
 
Last edited:
  • #20
Klaas van Aarsen said:
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:
 
  • #21
mathmari said:
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. 🤔
 
  • #22
Klaas van Aarsen said:
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:
 
  • #23
mathmari said:
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. o_O

Suppose we focus on the interval $[1,2]$.
Then $\ln t \ge \frac 12(t-1)$ isn't it? 🤔
 
  • #24
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:
 
  • #25
Theia said:
If \( t \approx 1 \), then \[ \ln t > 2 \frac{t-1}{t+1}. \]

Why does this inequality hold? :unsure:
Theia said:
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:
 
  • #26
Klaas van Aarsen said:
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:
 
  • #27
Klaas van Aarsen said:
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:
 
  • #28
mathmari said:
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:
\[ \ln x = 2\sum_{v=0}^{\infty} \frac{1}{2v+1} \left( \frac{x-1}{x+1} \right)^{2v+1} \]

mathmari said:
So we use this inequality in the interval $[0,1]$ or $[1,2]$ and we get divergence, right? :unsure:
Yes.
 
  • #29
Theia said:
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:
 
  • #30
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:
 

FAQ: Do These Integrals Converge or Diverge?

1. What is the purpose of checking convergence of integrals?

Checking convergence of integrals is important because it allows us to determine if the integral of a function is finite or infinite. This is useful in many areas of mathematics and science, such as in calculating probabilities, finding areas under curves, and solving differential equations.

2. How do you check the convergence of an integral?

There are several methods for checking the convergence of integrals, including the comparison test, the limit comparison test, the ratio test, and the root test. These tests involve comparing the given integral to a known integral or series with known convergence properties.

3. What is the difference between absolute convergence and conditional convergence?

Absolute convergence refers to an integral that converges regardless of the order in which the terms are added. On the other hand, conditional convergence refers to an integral that only converges when the terms are added in a specific order. This distinction is important when using certain convergence tests, such as the ratio test.

4. Can an integral converge at one point and diverge at another?

Yes, it is possible for an integral to converge at one point and diverge at another. This is known as a singular point or a point of discontinuity. In these cases, it is important to carefully analyze the behavior of the function near the singular point to determine the convergence properties of the integral.

5. Are there any real-world applications of checking convergence of integrals?

Yes, there are many real-world applications of checking convergence of integrals. For example, in physics, integrals are used to calculate the work done by a force, the center of mass of an object, and the energy of a system. In economics, integrals are used to calculate the area under a demand or supply curve, which represents the total economic value of a product or service. In engineering, integrals are used to calculate the total force or torque on a structure or system.

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