Integral tail and error
Hi,
I have an integral from 0 to ∞ and divided into two integrals, one from 0 to C and one from C to ∞. I'm trying to simplify the second integral so that I can solve for C to make the integral less than ε. [itex]\int\frac{1e^{x^{3}}}{x^{2}} = xf(x) + \int 3xe^{x^{3}} < \int 3xe^{x} = 3xe^{x} + \int 3e^{x} < \int 3e^{x} = 3e^{x} < ε[/itex] For ε = 1e^6 and limits from C to ∞ this gives me: [itex]3e^{C} < 10^{6} \Rightarrow C ≥ ln(3) + 6ln(10) ≈ 15[/itex]. However, Wolfram gives me that the value of the integral from 15 to ∞ is ≈0.0667 which is far from 1e^6. Is my inequality wrong? 
Re: Integral tail and error
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Difficult to answer if the term f(x) is not defined in the equation. And where the int 3x exp(3 x^3) is it comming from ? 
Re: Integral tail and error
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The primitive function of x^2 is x^1 then I use integration by parts on exp(x^3)*x^2 by letting h(x) = exp(x^3) and g'(x) = x^2 which gives h(x)g(x)  ∫h'(x)g(x)dx. h(x)g(x)*(x^1) is the term that I refer to as xf(x). 
Re: Integral tail and error
OK. It's clear.
3x exp(x^3) < 3x exp(x) is true, but the two functions are too much different. For example, case x=2.5 : 3x exp(x^3)=0.00000123 while 3x exp(x)=0.616 So, it is not surpristng that the result of the respective integrals be very far from one to the other. I suggest 3x exp(x^3) < 3x² exp(x^3) Integral (from x=C to infinity) of 3x²exp(x^3) dx = exp(C^3)= 10^(6) Leading to C= 2.4 Integral (from x=2.4 to infinity) of 3x exp(x^3) dx = 4*10^(7) approximately 
Re: Integral tail and error
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Good suggestion, didn't think of it... remember though that the original function, f(x) is (1  exp(x^3))/x^2. The integral of that one from 2.4 to infinity is approx. 0.417 according to WolframAlpha. But that's what I don't understand, how can the integral of the original function be larger then that of the approximated function (which is larger)? 0.417 is much larger then 10^7? http://goo.gl/1TcPa (you have to put a parenthesis around 1  exp(...)) 
Re: Integral tail and error
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The tail of this term 1/x² is : Integal (from x=2.4 to infinity) of dx/x² = 1/2.4 = 0.417 With C=2.4, the tail of the term exp(x^3)/x² is what was said in my previous post : less than 10^7. If you consider the whole function (1  exp(x^3))/x^2 , it is completely different : You must not integrate 1/x² separately. Start with (1  exp(x^3))/x^2 < 1/x² Integral (x=C to infinity) of 1/x² = 1/C = 10^6 C= 10^6 Integral (from x=10^6 to infinity) of ((1  exp(x^3))/x^2)dx = 10^6 approximately 
Re: Integral tail and error
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Thanks! 
Re: Integral tail and error
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Of course, you are able to calculate separately the integrals and add them up. But, also you are able to calculate separately the tail of each of them and add them up. Not only one of the two tails has to be < 106 , but the sum of the two. Second : On another viewpoint, you could say : I know exactly the integral of 1/x², which is 1/x. So there is no error on this part. As a consequence, I am interested only about the error on the integral of (exp(x^3))/x. It is completely different : The tail considered is only the tail of the integral of (exp(x^3))/x. As already shown, C=2.4 and this is to be tested NOT with the whole function (1exp(x^3))/x but on the part (exp(x^3))/x .The integral (from x=2.4 to infinity) of (exp(x^3))/x is < 10^6 as expected. 
Re: Integral tail and error
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