Why does a u-sub go wrong when finding values for q such that $int_{1}^infty x^q e^{x^{q+1}} dx$ converges or...
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Why does a u-sub go wrong when finding values for q such that $int_{1}^infty x^q e^{x^{q+1}} dx$ converges or diverges?
The Next CEO of Stack OverflowCalculate $int_{0}^{infty}frac{arctan(x)}{(1+x^2)^{3/2}}dx$ or show that it divergesFind the Values of $p$ and $q$ such that $int_0^{infty} x^pln(1+x)^q dx$ Converges or Diverges.Show that $int_{0}^{1}x^pdx$ converges for $p>-1$ and diverges otherwise.Analysis: Prove that this improper integral diverges but the limit as $t rightarrow infty=pi$?How to determine whether an improper integral converges or diverges?Prove that $int_0^{infty} frac{sin x}{x^p}, dx$ converges for $0<p<2$integral $begin{equation} int limits_{0}^{infty} frac{e^{-sqrt{x}}}{sqrt{x}} dx end{equation}$Why the integral converges only when C is 1 and why other values of C would give infinity?For what values of ''$a$'' , does this Improper Integral converges?Finding all values $p$ for which $int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$ converges
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I'm trying to determine for which values of $qinmathbb{R}$ this improper integral converges and diverges. I have discrepancy in my solutions, so there appears to be something wrong, or some subtle nuance I'm unaware of.
$$int_{1}^infty x^q e^{x^{q+1}} dx$$
Let $u=x^{q+1}$, so $frac{du}{dx}=(q+1)e^q$. Then,
$$frac{1}{q+1}int_{1}^infty e^u du=frac{1}{q+1} lim_{Lrightarrow infty}int_{1}^L e^u du= frac{1}{q+1}lim_{Lrightarrow infty} [e^u]_{1}^L$$
Now if I sub $L$ and $1$ in, I get divergence to $infty$ if $q>-1$ and divergence to $-infty$ if $q<-1$. However, if I sub $x^{q+1}$ back in for $u$, I get,
$$frac{1}{q+1}lim_{Lrightarrow infty} [e^{x^{q+1}}]_{1}^L = frac{1}{q+1} (lim_{Lrightarrow infty} (e^{L^{q+1}}) - e)$$
Hence, the integral diverges to $infty$ for $qgeq-1$ and converges to $frac{1-e}{q+1}$ if $q<-1$.
Why do the solutions change if I sub $x^{q+1}$ back in? Which solution set is correct and why?
integration convergence improper-integrals
edited Mar 16 at 18:45
Robert Howard
2,2783935
asked May 20 '18 at 18:02
HumptyDumptyHumptyDumpty
358110
$endgroup$
add a comment |
$begingroup$
I'm trying to determine for which values of $qinmathbb{R}$ this improper integral converges and diverges. I have discrepancy in my solutions, so there appears to be something wrong, or some subtle nuance I'm unaware of.
$$int_{1}^infty x^q e^{x^{q+1}} dx$$
Let $u=x^{q+1}$, so $frac{du}{dx}=(q+1)e^q$. Then,
$$frac{1}{q+1}int_{1}^infty e^u du=frac{1}{q+1} lim_{Lrightarrow infty}int_{1}^L e^u du= frac{1}{q+1}lim_{Lrightarrow infty} [e^u]_{1}^L$$
Now if I sub $L$ and $1$ in, I get divergence to $infty$ if $q>-1$ and divergence to $-infty$ if $q<-1$. However, if I sub $x^{q+1}$ back in for $u$, I get,
$$frac{1}{q+1}lim_{Lrightarrow infty} [e^{x^{q+1}}]_{1}^L = frac{1}{q+1} (lim_{Lrightarrow infty} (e^{L^{q+1}}) - e)$$
Hence, the integral diverges to $infty$ for $qgeq-1$ and converges to $frac{1-e}{q+1}$ if $q<-1$.
Why do the solutions change if I sub $x^{q+1}$ back in? Which solution set is correct and why?
integration convergence improper-integrals
edited Mar 16 at 18:45
Robert Howard
2,2783935
asked May 20 '18 at 18:02
HumptyDumptyHumptyDumpty
358110
$endgroup$
2
$begingroup$
Are you sure that $uto+infty$ (your upper integration bound) as $xto+infty$ when $q<-1$?
$endgroup$
– A.Γ.
May 20 '18 at 18:14
$begingroup$
@A.Γ. Thanks, that answers my question :)
$endgroup$
– HumptyDumpty
May 20 '18 at 18:17
$begingroup$
@A.Γ. Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems like it was exactly the kind of advice OP was looking for.
$endgroup$
– Robert Howard
Dec 1 '18 at 2:07
add a comment |
$begingroup$
I'm trying to determine for which values of $qinmathbb{R}$ this improper integral converges and diverges. I have discrepancy in my solutions, so there appears to be something wrong, or some subtle nuance I'm unaware of.
$$int_{1}^infty x^q e^{x^{q+1}} dx$$
Let $u=x^{q+1}$, so $frac{du}{dx}=(q+1)e^q$. Then,
$$frac{1}{q+1}int_{1}^infty e^u du=frac{1}{q+1} lim_{Lrightarrow infty}int_{1}^L e^u du= frac{1}{q+1}lim_{Lrightarrow infty} [e^u]_{1}^L$$
Now if I sub $L$ and $1$ in, I get divergence to $infty$ if $q>-1$ and divergence to $-infty$ if $q<-1$. However, if I sub $x^{q+1}$ back in for $u$, I get,
$$frac{1}{q+1}lim_{Lrightarrow infty} [e^{x^{q+1}}]_{1}^L = frac{1}{q+1} (lim_{Lrightarrow infty} (e^{L^{q+1}}) - e)$$
Hence, the integral diverges to $infty$ for $qgeq-1$ and converges to $frac{1-e}{q+1}$ if $q<-1$.
Why do the solutions change if I sub $x^{q+1}$ back in? Which solution set is correct and why?
integration convergence improper-integrals
edited Mar 16 at 18:45
Robert Howard
2,2783935
asked May 20 '18 at 18:02
HumptyDumptyHumptyDumpty
358110
$endgroup$
I'm trying to determine for which values of $qinmathbb{R}$ this improper integral converges and diverges. I have discrepancy in my solutions, so there appears to be something wrong, or some subtle nuance I'm unaware of.
$$int_{1}^infty x^q e^{x^{q+1}} dx$$
Let $u=x^{q+1}$, so $frac{du}{dx}=(q+1)e^q$. Then,
$$frac{1}{q+1}int_{1}^infty e^u du=frac{1}{q+1} lim_{Lrightarrow infty}int_{1}^L e^u du= frac{1}{q+1}lim_{Lrightarrow infty} [e^u]_{1}^L$$
Now if I sub $L$ and $1$ in, I get divergence to $infty$ if $q>-1$ and divergence to $-infty$ if $q<-1$. However, if I sub $x^{q+1}$ back in for $u$, I get,
$$frac{1}{q+1}lim_{Lrightarrow infty} [e^{x^{q+1}}]_{1}^L = frac{1}{q+1} (lim_{Lrightarrow infty} (e^{L^{q+1}}) - e)$$
Hence, the integral diverges to $infty$ for $qgeq-1$ and converges to $frac{1-e}{q+1}$ if $q<-1$.
Why do the solutions change if I sub $x^{q+1}$ back in? Which solution set is correct and why?
integration convergence improper-integrals
integration convergence improper-integrals
edited Mar 16 at 18:45
Robert Howard
2,2783935
asked May 20 '18 at 18:02
HumptyDumptyHumptyDumpty
358110
edited Mar 16 at 18:45
Robert Howard
2,2783935
asked May 20 '18 at 18:02
HumptyDumptyHumptyDumpty
358110
edited Mar 16 at 18:45
Robert Howard
2,2783935
edited Mar 16 at 18:45
Robert Howard
2,2783935
edited Mar 16 at 18:45
Robert Howard
2,2783935
2,2783935
asked May 20 '18 at 18:02
HumptyDumptyHumptyDumpty
358110
asked May 20 '18 at 18:02
HumptyDumptyHumptyDumpty
358110
asked May 20 '18 at 18:02
HumptyDumptyHumptyDumpty
358110
358110
2
$begingroup$
Are you sure that $uto+infty$ (your upper integration bound) as $xto+infty$ when $q<-1$?
$endgroup$
– A.Γ.
May 20 '18 at 18:14
$begingroup$
@A.Γ. Thanks, that answers my question :)
$endgroup$
– HumptyDumpty
May 20 '18 at 18:17
$begingroup$
@A.Γ. Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems like it was exactly the kind of advice OP was looking for.
$endgroup$
– Robert Howard
Dec 1 '18 at 2:07
add a comment |
2
$begingroup$
Are you sure that $uto+infty$ (your upper integration bound) as $xto+infty$ when $q<-1$?
$endgroup$
– A.Γ.
May 20 '18 at 18:14
$begingroup$
@A.Γ. Thanks, that answers my question :)
$endgroup$
– HumptyDumpty
May 20 '18 at 18:17
$begingroup$
@A.Γ. Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems like it was exactly the kind of advice OP was looking for.
$endgroup$
– Robert Howard
Dec 1 '18 at 2:07
2
2
$begingroup$
Are you sure that $uto+infty$ (your upper integration bound) as $xto+infty$ when $q<-1$?
$endgroup$
– A.Γ.
May 20 '18 at 18:14
$begingroup$
Are you sure that $uto+infty$ (your upper integration bound) as $xto+infty$ when $q<-1$?
$endgroup$
– A.Γ.
May 20 '18 at 18:14
$begingroup$
@A.Γ. Thanks, that answers my question :)
$endgroup$
– HumptyDumpty
May 20 '18 at 18:17
$begingroup$
@A.Γ. Thanks, that answers my question :)
$endgroup$
– HumptyDumpty
May 20 '18 at 18:17
$begingroup$
@A.Γ. Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems like it was exactly the kind of advice OP was looking for.
$endgroup$
– Robert Howard
Dec 1 '18 at 2:07
$begingroup$
@A.Γ. Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems like it was exactly the kind of advice OP was looking for.
$endgroup$
– Robert Howard
Dec 1 '18 at 2:07
add a comment |
0
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$begingroup$
Are you sure that $uto+infty$ (your upper integration bound) as $xto+infty$ when $q<-1$?
$endgroup$
– A.Γ.
May 20 '18 at 18:14
$begingroup$
@A.Γ. Thanks, that answers my question :)
$endgroup$
– HumptyDumpty
May 20 '18 at 18:17
$begingroup$
@A.Γ. Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems like it was exactly the kind of advice OP was looking for.
$endgroup$
– Robert Howard
Dec 1 '18 at 2:07