1. Marine, integral question

Hi!

$\displaystyle \int_0^{\infty}\frac{dx}{x^2+e^{-x}}$

(the assignment actually requires to prove if the integral converges on $\displaystyle (0,\infty)$

thanks!

marine
We have

$\displaystyle \int_0^{\infty}\frac{dx}{x^2+e^{-x}}$

Well we know that

$\displaystyle \forall{x}>1\quad{0\leq\frac{1}{x^2+e^{-x}}\leq\frac{1}{x^2}}$

This implies that

$\displaystyle \quad\int_1^{\infty}0~dx\leq\int_1^{\infty}\frac{d x}{x^2+e^{-x}}\leq\int_1^{\infty}\frac{dx}{x^2}$

We can see that if we evaluate the left and right hand integrals we get

$\displaystyle 0\leq\int_1^{\infty}\frac{dx}{x^2+e^{-x}}\leq{1}$

and secondly and this is up to you to prove

$\displaystyle \int_0^1\frac{dx}{x^2+e^{-x}}$ is a finite value which we will denote $\displaystyle C$

We see that

$\displaystyle C\leq{C+\int_1^{\infty}\frac{dx}{x^2+e^{-x}}}\leq{C+1}$

$\displaystyle \Rightarrow\quad{C\leq\int_0^{\infty}\frac{dx}{x^2 +e^{-x}}\leq{C+1}}$

This implies that this integral is bounded by two finite values, therefore convergent.

2. Originally Posted by Mathstud28
We have

$\displaystyle \int_0^{\infty}\frac{dx}{x^2+e^{-x}}$

Well we know that

$\displaystyle \forall{x}>1\quad{0\leq\frac{1}{x^2+e^{-x}}\leq\frac{1}{x^2}}$

This implies that

$\displaystyle \quad\int_1^{\infty}0~dx\leq\int_1^{\infty}\frac{d x}{x^2+e^{-x}}\leq\int_1^{\infty}\frac{dx}{x^2}$

We can see that if we evaluate the left and right hand integrals we get

$\displaystyle 0\leq\int_1^{\infty}\frac{dx}{x^2+e^{-x}}\leq{1}$

and secondly and this is up to you to prove

$\displaystyle \in_0^1\frac{dx}{x^2+e^{-x}}$ is a finite value which we will denote $\displaystyle C$

We see that

$\displaystyle C\leq{C+\int_1^{\infty}\frac{dx}{x^2+e^{-x}}\leq{C+1}$

$\displaystyle \Rightarrow\quad{C\leq\int_0^{\infty}\frac{dx}{x^2 +e^{-x}}\leq{C+1}$

This implies that this integral is bounded by two finite values, therefore convergent.
thanks, Mathstud28, for helping I almost got it. Unfortunately the last 2 lines I cannot read It says 'LaTex error' . Could you try posting them again?

Btw, by this method we did our task (actually you - mine ) but this does not give the exact limit the integral converges to.

What if we had to find it's value? How should we proceed?

3. Originally Posted by Marine
thanks, Mathstud28, for helping I almost got it. Unfortunately the last 2 lines I cannot read It says 'LaTex error' . Could you try posting them again?

Btw, by this method we did our task (actually you - mine ) but this does not give the exact limit the integral converges to.

What if we had to find it's value? How should we proceed?
It has no nice number representation. It is approximated by 1.986

4. I feel like doing this problem in full for some reason.

Show that the following integral converges.

$\displaystyle \int_0^{\infty}\frac{dx}{x^2+e^{-x}}~dx=I$

The first part to this is to rewrite it as

$\displaystyle \int_0^1\frac{dx}{x^2+e^{-x}}+\int_1^{\infty}\frac{dx}{x^2+e^{-x}}~dx$

Now let $\displaystyle I_1=\int_0^1\frac{dx}{x^2+e^{-x}}~dx$

and

$\displaystyle I_2=\int_1^{\infty}\frac{dx}{x^2+e^{-x}}~dx$

Now we can see that $\displaystyle I_1+I_2=I$

Now first let us consider the convergence of $\displaystyle I_2$

We know that

$\displaystyle \forall{x}\geq{1}\quad0\leq\frac{1}{x^2+e^{-x}}\leq\frac{1}{x^2}$

Now if we define $\displaystyle D=[1,\infty)$

and $\displaystyle R=\left\{x:x\geq{1}\right\}$

We see that $\displaystyle D=R$

and noting that

$\displaystyle \int_1^{\infty}f(x)=\int_{R}f(x)$

we can conclude that

$\displaystyle \int_1^{\infty}0~dx\leq{I_2}\leq\int_1^{\infty}\fr ac{dx}{x^2}$

$\displaystyle \Rightarrow{0\leq{I_2}\leq{1}}$

Now we must consider $\displaystyle I_1$

We know that

$\displaystyle \forall{x}\in[0,1]\quad\frac{1}{x^2+1}\leq\frac{1}{x^2+e^{-x}}\leq{\frac{1}{e^{-x}}}=e^x$

So now once again seeing that

if we our region of integration is included in we may make the next assumption that

$\displaystyle \int_0^1\frac{dx}{x^2+1}\leq{I_1}\leq{\int_0^1{e^x }~dx}$

$\displaystyle \Rightarrow\quad\frac{\pi}{4}\leq{I_1}\leq{e-1}$

Now let us denote $\displaystyle I_1=C\in\left[\frac{\pi}{4},e-1\right]$

Then we can finally conclude that

$\displaystyle C\in\left[\frac{\pi}{4},e-1\right]\leq{I_1+I_2}\leq{C\in\left[\frac{\pi}{4},e-1\right]}$

$\displaystyle \Rightarrow\quad{C\in\left[\frac{\pi}{4},e-1\right]}\leq{I}\leq{C\in\left[\frac{\pi}{4},e-1\right]+1}$

This implies that $\displaystyle I$ is bounded by two finite values.

$\displaystyle \therefore\quad{I=\int_0^{\infty}\frac{dx}{x^2+e^{-x}}~dx}\quad\text{is convergent}\quad\blacksquare$

5. now this is like a book-written solution Thanks a lot for it

since I have not used the method before I would like to ask a few more questions :P

1. The last step was (in a more generalised statement)

if a, and b are constants (or some constant intervalls) and a<b, then the integal I(f(x)) of the function f(x) converges, if:

a < I(f(x)) < b

question: if f(x) = sinx, I(sinx) = -cosx from 0 to some point in infinity, a = -2 and b = 2 the inequality:
-2 </= I(sinx) </= 2 ist true, but sinx does not converge. Why?

2. We carry back to my first problem, again the last step:

we had: integrals I_1 on [a;b] and I_2 on [c;d]

On which intervalls do we define the operations:
I_1+I_2
I_1*I_2
I_1^(I_2)

do se take the intersections or just perform the operations seperately for each pair of limits? (I assume the inverse operations will obey the same rules, if wring, correct me)

Thnaks again for the help,

best wishes, Marine

6. Originally Posted by Marine
now this is like a book-written solution Thanks a lot for it

since I have not used the method before I would like to ask a few more questions :P

1. The last step was (in a more generalised statement)

if a, and b are constants (or some constant intervalls) and a<b, then the integal I(f(x)) of the function f(x) converges, if:

a < I(f(x)) < b

question: if f(x) = sinx, I(sinx) = -cosx from 0 to some point in infinity, a = -2 and b = 2 the inequality:
-2 </= I(sinx) </= 2 ist true, but sinx does not converge. Why?

2. We carry back to my first problem, again the last step:

we had: integrals I_1 on [a;b] and I_2 on [c;d]

On which intervalls do we define the operations:
I_1+I_2
I_1*I_2
I_1^(I_2)

do se take the intersections or just perform the operations seperately for each pair of limits? (I assume the inverse operations will obey the same rules, if wring, correct me)

Thnaks again for the help,

best wishes, Marine

You have this I think

$\displaystyle -1\leq{\sin(x)}\leq{1}\quad\forall{x}\in\mathbb{R}$

$\displaystyle \int_0^{\infty}\sin(x)~dx$ does not converge? Well consdier that neither of your bounding functions (-1 and 1) integrals from 0 to infinity are convergent =P. So this tells us nothing.

Consider this

$\displaystyle \forall{x}\geq{-1}\quad{-1\leq\frac{1}{x}\leq{1}}$

Now evalutiating the integrals gives us that

$\displaystyle -\infty\leq\int_1^{\infty}\frac{dx}{x}\leq\infty$

This tells us nothing since our integral could be a lower order infinity etc.

As for yoru second question I belive the answer would be this?

$\displaystyle \int_a^{b}f(x)~dx+\int_b^{c}f(x)~dx=\int_a^{c}f(x) ~dx$

Assuming $\displaystyle f(x)$ is continous on that domain

7. Originally Posted by Mathstud28
You have this I think

$\displaystyle -1\leq{\sin(x)}\leq{1}\quad\forall{x}\in\mathbb{R}$

$\displaystyle \int_0^{\infty}\sin(x)~dx$ does not converge? Well consdier that neither of your bounding functions (-1 and 1) integrals from 0 to infinity are convergent =P. So this tells us nothing.

Consider this

$\displaystyle \forall{x}\geq{-1}\quad{-1\leq\frac{1}{x}\leq\1}$

Now evalutiating the integrals gives us that

$\displaystyle -\infty\leq\int_1^{\infty}\frac{dx}{x}\leq\infty$

This tells us nothing since our integral could be a lower order infinity etc.

As for yoru second question I belive the answer would be this?

$\displaystyle \int_a^{b}f(x)~dx+\int_b^{c}f(x)~dx=\int_a^{c}f(x) ~dx$

Assuming $\displaystyle f(x)$ is continous on that domain
1. I meant the area under the sine curve is strongly restricted from -2 to 2 from 0 to any point in +oo, cuz: Int sinx dx[0 do pi] = 2 and Int sinx dx[pi to 2pi] = -2, so the two repeating parts cancel each other out during the progress to +oo

2. I_1 lies between [pi/4; e-1] and I_2 between [0 ; 1]. Where does I_1+I_2 lie?

[pi/4+0 ; e-1+1] - as I think you did, or the intersections: [pi/1 ; 1]?

Sorry but I still can't figure out how you did the last step

8. Originally Posted by Marine
1. I meant the area under the sine curve is strongly restricted from -2 to 2 from 0 to any point in +oo, cuz: Int sinx dx[0 do pi] = 2 and Int sinx dx[pi to 2pi] = -2, so the two repeating parts cancel each other out during the progress to +oo

2. I_1 lies between [pi/4; e-1] and I_2 between [0 ; 1]. Where does I_1+I_2 lie?

[pi/4+0 ; e-1+1] - as I think you did, or the intersections: [pi/1 ; 1]?

Sorry but I still can't figure out how you did the last step
Haha, well your first question is actually dealing with a subject beyond me as of now, namely Fourier Analysis and Complex Analysis.

As for your second all I did was add the two inequalities

I can add a number to each part of an inequality right?

So I just added I_1 to each part but on the left and right I just called it C which has numerical bounds

9. 'Haha, well your first question is actually dealing with a subject beyond me as of now, namely Fourier Analysis and Complex Analysis.'

I've watched a few movies on Fourier-Series and Fourier-Transform and cannot imagine you solve such complex integrals and Fourier Analysis is 'a subject beyond' you

Do you do this complexified calculus stuff at school or you read it by yourself?

10. Originally Posted by Marine
'Haha, well your first question is actually dealing with a subject beyond me as of now, namely Fourier Analysis and Complex Analysis.'

I've watched a few movies on Fourier-Series and Fourier-Transform and cannot imagine you solve such complex integrals and Fourier Analysis is 'a subject beyond' you

Do you do this complexified calculus stuff at school or you read it by yourself?
By myself, my school does not offer my level of mathematics (I will be a senior in highschool next year)

11. well done!

so if you have some time, and find it interesting, you could check this threads

(that stuff by donylee)

this is where I got familiar with these topics - they pretty much use school calculus skills to build upon

12. Originally Posted by Marine
well done!

so if you have some time, and find it interesting, you could check this threads

(that stuff by donylee)

this is where I got familiar with these topics - they pretty much use school calculus skills to build upon
Yeah its cool stuff, but I think I will wait until I get to it. If you try to learn someting before you have the prerequisite skills you often end up misunderstanding half of the pertinent material. Thanks for the links by the way.

13. 'If you try to learn something before you have the prerequisite skills you often end up misunderstanding half of the pertinent material.'

I agree, but haven't you taken this risk before?! - actually I got the impression you're ready from a 'prerequisite skills' point of view - but it also depends (and I think that's the decisive part) on the tutor - anyway you can check it, if you get bored

Do you know such links or scripts, suitable for an amateur like me for double and triple integrals and maybe these techniques you use when solving more complex integrals?

14. Originally Posted by Marine
'If you try to learn something before you have the prerequisite skills you often end up misunderstanding half of the pertinent material.'

I agree, but haven't you taken this risk before?! - actually I got the impression you're ready from a 'prerequisite skills' point of view - but it also depends (and I think that's the decisive part) on the tutor - anyway you can check it, if you get bored

Do you know such links or scripts, suitable for an amateur like me for double and triple integrals and maybe these techniques you use when solving more complex integrals?
Honestly, if you want to learn about stuff like that. Just stick around here and stay low. Read the threads and absorb.