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Improper integral question

  1. Jan 24, 2008 #1
    Last edited: Jan 24, 2008
  2. jcsd
  3. Jan 24, 2008 #2
    [tex]\int_0^{\infty}xe^{-x^2}dx[/tex]

    [tex]u=-x^2dx[/tex]
    [tex]-\frac 1 2 du=xdx[/tex]

    [tex]-\frac 1 2 \int_0^{\infty}e^u du[/tex]

    [tex]-\frac 1 2 e^{-x^2} |_0^{\infty}[/tex]

    [tex]-\frac{1}{2e^{\infty^2}}-\left(-\frac{1}{2e^0}\right)[/tex]

    It's not always a good idea to evaluate for the limit as your substituting.
     
    Last edited: Jan 24, 2008
  4. Jan 24, 2008 #3
    where is my mistake in they way i've solved??

    i my way didnt even came to the limit part

    i was told to use limit for the interval that has infinity in it
     
  5. Jan 24, 2008 #4
    You switched your limits while substituting. Do it again using 1 to infinity.
     
  6. Jan 24, 2008 #5
    i need to switch the intervals if i want change my variables (in order to solve the question)

    thats what i did
    i dont see any reason why i need to change my variables
    again for having the old intervals??

    or what you are trying to say is
    "that i can change variables in a way that keeps the old intervals"

    in that case its impossible to do because every
    subtitution changes the intervals
     
    Last edited: Jan 24, 2008
  7. Jan 24, 2008 #6
    why is my solution wronge??
     
    Last edited: Jan 24, 2008
  8. Jan 24, 2008 #7
    You made two mistakes. The first one is that you switched the limits after substituting, the integral should read:

    [tex]\int_{1}^{0}\frac{t}{-2}dt[/tex]

    The second one is that you do not need to inverse substitute the function. Why did you do this? You can integrate it directly with t as variable. Look back to the meaning of the substitution for definite integrals in your textbook. In case of an integral without limits, you should do this. You can do it here, but you will make it only more difficult on yourself.
     
  9. Jan 24, 2008 #8
    i switched the intervals because
    the higher value always need to be on the top.

    even then the "1" interval gives me a different answer the the book does
    ???????

    regrading your second remark that you sayd about reverse substitution
    thow its a long way
    i didnt do anything wronge there
     
  10. Jan 24, 2008 #9
    well as others pointed out, here is what u did wrong that u did not get the correct result:
    u took this substitution [tex] t=e^{-x^{2}}[/tex],
    now after u defferentiate we get [tex] -2xe^{-x^{2}}dx=dt[/tex]or
    [tex]-2xtdx=dt[/tex] now devide by -2t and we get,( you should have done this way)
    [tex]xdx=\frac{dt}{-2t}[/tex] and now go back and substitute for xdx, and also for [tex] t=e^{-x^{2}}[/tex]

    [tex]\int_0^{\infty}xe^{-x^2}dx=\frac{-1}{2}\int_1^{0}\frac{tdt}{t}=-\frac{-1}{2}\int_0^{1}dt=\frac{1}{2}\int_0^{1}dt[/tex]and here is your second mistake,you did not change the sign of the ingegral when you switched boundaries.
    Now i am sure u know how to go about the rest, so the answer obviously will be 1/2.
    Look i am not even sure that this is a safe way to pursue, the reason that we got to the correct result might have been accidental, look it is more safe to use limits in these cases, at least i prefere to use them.
     
  11. Jan 25, 2008 #10
    in the end i got to the same integral as you showed

    regarding my second mistake:
    that i forgot to multiply by a minus


    but still i get the same "e" type expression
    because as you showed ,the antiderivative will be
    (0.5)t from one to zero
    now when we substitute t with the original expression we get "e" type expression
    and afiter solveing it the solution is:
    1/(2*e) - 1/2

    not 1/2
     
  12. Jan 25, 2008 #11

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    First, it is NOT true that "the higher value always need to be on the top." That is simply a convention for convenience. It is true that [itex]\int_a^b f dx= -\int_b^a f dx[/itex]. I you want to swap limits of integration you must also multiply by -1.

    You did, correctly, substitute [itex]t= e^{-x^2}[/itex], change the limits of integration from x= 0, x=[itex]\infty[/itex] to t= 1, t= 0. You correctly integrated to get
    [itex](-1/2)t]_{t=1}^0[/itex].

    Unfortunately, you then changed the function back to "x" and evaluated at the "t" values! Either
    [tex](-1/2)t]_{t=1}^0= [(-1/2)(0)-(-1/2)1]= 1/2[/tex]
    or
    [tex](-1/2)e^{-x^2}]_{x= 0}^\infty= [(-1/2)(0)-(-1/2)(1)]= 1/2[/tex]
    would be correct.
     
  13. Jan 25, 2008 #12
    ooooohhhhhhhhhhhhhhh

    i am such an idiot you are totaly right
    finaly i got my redemption

    thanks
     
  14. Jan 25, 2008 #13
    Yeah, one great thing about definite integrals is that there is no need to go back to the original variable after a couple of substitutions that you might have made,however, if you really love so much the original variable, then you have to keep track of all the change in limits of integration that you have made on your way, and sometimes this is painful!
    take care!
     
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