Calculate the antiderivatives

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  • #1
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Homework Statement


[itex]\int \frac{ln(x^{2}+4^{x})}{\sqrt{3x^{7}+7x^{^3}}}dx[/itex]


Homework Equations


X.


The Attempt at a Solution


Wolfram Alpha seem to give no answer.
 
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Answers and Replies

  • #2
tiny-tim
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the x2 + 4x makes that impossible to do by analytic means
 
  • #3
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Actually my original problem was determining the convergence or divergence of the following improper integral:
[itex]\int^{+∞}_{0} \frac{ln(x^{2}+4^{x})}{\sqrt{3x^{7}+7x^{^3}}}dx[/itex]
I split the integral into
[itex]\int^{+∞}_{1} \frac{ln(x^{2}+4^{x})}{\sqrt{3x^{7}+7x^{^3}}}dx[/itex]
and [itex]\int^{1}_{0} \frac{ln(x^{2}+4^{x})}{\sqrt{3x^{7}+7x^{^3}}}dx[/itex]
, calculate the antiderivatives, then evaluate the limit of them.
So if I can't calculate the antiderivative, is there any alternative way to see if this integral convergent or not ??
 
  • #4
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Tiny-Tim, how did you know thats impossible to do by analytical means? What should I google to learn more?
 
  • #5
2,967
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Actually my original problem was determining the convergence or divergence of the following improper integral:
[itex]\int^{+∞}_{0} \frac{ln(x^{2}+4^{x})}{\sqrt{3x^{7}+7x^{^3}}}dx[/itex]
I split the integral into
[itex]\int^{+∞}_{1} \frac{ln(x^{2}+4^{x})}{\sqrt{3x^{7}+7x^{^3}}}dx[/itex]
and [itex]\int^{1}_{0} \frac{ln(x^{2}+4^{x})}{\sqrt{3x^{7}+7x^{^3}}}dx[/itex]
, calculate the antiderivatives, then evaluate the limit of them.
So if I can't calculate the antiderivative, is there any alternative way to see if this integral convergent or not ??

Do you know how to find the asymptotic behavior of your integrand for [itex]x \rightarrow \infty[/itex] and [itex]x \rightarrow 0[/itex]? If yes, then you may use the comparison test.
 
  • #6
tiny-tim
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hi superg33k! :smile:
Tiny-Tim, how did you know thats impossible to do by analytical means? What should I google to learn more?

it's obvious just from looking at it … that bracket is simply too complicated for any of the known techniques to work! :redface:

once you've had lots of practice at differentiating and integrating, you'll see why :smile:
 
  • #7
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Do you know how to find the asymptotic behavior of your integrand for [itex]x \rightarrow \infty[/itex] and [itex]x \rightarrow 0[/itex]? If yes, then you may use the comparison test.

I'll start off. Let us consider the upper bound first. For [itex]x \rightarrow \infty[/itex]. [itex]x^2 = o(4^x)[/itex], so the logarithm in the numerator behaves as [itex]\sim x \, \ln(4)[/itex]. Similarly, [itex]7 x^3 = o(3 x^7)[/itex], so the expression under the square root in the denominator behaves as [itex]\sim 3 x^7[/itex]. Therefore, the integrand behaves as:
[tex]
\sim \frac{x \, \ln(4)}{\sqrt{3 x^7}} = \frac{\ln(4)}{\sqrt{3}} \, x^{-5/2}
[/tex]
Do you know whether the integral:
[tex]
\int_{1}^{\infty}{x^{-5/2} \, dx}
[/tex]
is convergent or divergent?

A similar analysis can be done on the lower bound of the integral. However, what are the dominant terms in this limit?
 
  • #8
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Thank you very much, I can take it from here :D.
 

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