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Math Help - Marine, integral question

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Marine, integral question

    Hi!

    Can someone please help me solving this improper integral:

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

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


    thanks!

    marine
    We have

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


    Well we know that

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

    This implies that

    \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

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

    and secondly and this is up to you to prove

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

    We see that

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

    \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.
    Last edited by Mathstud28; July 19th 2008 at 10:08 AM.
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  2. #2
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    Quote Originally Posted by Mathstud28 View Post
    We have

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


    Well we know that

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

    This implies that

    \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

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

    and secondly and this is up to you to prove

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

    We see that

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

    \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?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Marine View Post
    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
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    I feel like doing this problem in full for some reason.

    Show that the following integral converges.

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

    The first part to this is to rewrite it as

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

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

    and

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

    Now we can see that I_1+I_2=I


    Now first let us consider the convergence of I_2

    We know that

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

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

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

    We see that D=R

    and noting that

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

    we can conclude that

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


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


    Now we must consider I_1

    We know that

    \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

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


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


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

    Then we can finally conclude that

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


    \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 I is bounded by two finite values.

    \therefore\quad{I=\int_0^{\infty}\frac{dx}{x^2+e^{-x}}~dx}\quad\text{is convergent}\quad\blacksquare
    Last edited by Mathstud28; July 19th 2008 at 01:23 PM. Reason: typo
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  5. #5
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    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
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Marine View Post
    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

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

    So your asking why

    \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

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

    Now evalutiating the integrals gives us that

    -\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?

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

    Assuming f(x) is continous on that domain
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    You have this I think

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

    So your asking why

    \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

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

    Now evalutiating the integrals gives us that

    -\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?

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

    Assuming 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
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Marine View Post
    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
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  9. #9
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    '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?
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Marine View Post
    '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)
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  11. #11
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    well done!

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

    ODEs: YouTube - Broadcast Yourself.

    Fourier Series: YouTube - Broadcast Yourself.

    (that stuff by donylee)


    this is where I got familiar with these topics - they pretty much use school calculus skills to build upon
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Marine View Post
    well done!

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

    ODEs: YouTube - Broadcast Yourself.

    Fourier Series: YouTube - Broadcast Yourself.

    (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.
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    '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?
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Marine View Post
    '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.
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