Comparing Improper Integrals Involving Infinity
$begingroup$
From my current understanding: $K>J$ and $L>K$ , therefore $L>K>J$. How can I compare the first integral $I$ ?
calculus integration convergence infinity
edited Dec 23 '18 at 20:29
Glorfindel
3,41981830
asked Jan 16 '13 at 3:26
EngGenieEngGenie
12019
$endgroup$
add a comment |
$begingroup$
From my current understanding: $K>J$ and $L>K$ , therefore $L>K>J$. How can I compare the first integral $I$ ?
calculus integration convergence infinity
edited Dec 23 '18 at 20:29
Glorfindel
3,41981830
asked Jan 16 '13 at 3:26
EngGenieEngGenie
12019
$endgroup$
1
$begingroup$
Hint: $x < e^x$ so $I < ldots$.
$endgroup$
– Robert Israel
Jan 16 '13 at 4:12
add a comment |
$begingroup$
From my current understanding: $K>J$ and $L>K$ , therefore $L>K>J$. How can I compare the first integral $I$ ?
calculus integration convergence infinity
edited Dec 23 '18 at 20:29
Glorfindel
3,41981830
asked Jan 16 '13 at 3:26
EngGenieEngGenie
12019
$endgroup$
From my current understanding: $K>J$ and $L>K$ , therefore $L>K>J$. How can I compare the first integral $I$ ?
calculus integration convergence infinity
calculus integration convergence infinity
edited Dec 23 '18 at 20:29
Glorfindel
3,41981830
asked Jan 16 '13 at 3:26
EngGenieEngGenie
12019
edited Dec 23 '18 at 20:29
Glorfindel
3,41981830
asked Jan 16 '13 at 3:26
EngGenieEngGenie
12019
edited Dec 23 '18 at 20:29
Glorfindel
3,41981830
edited Dec 23 '18 at 20:29
Glorfindel
3,41981830
edited Dec 23 '18 at 20:29
Glorfindel
3,41981830
3,41981830
asked Jan 16 '13 at 3:26
EngGenieEngGenie
12019
asked Jan 16 '13 at 3:26
EngGenieEngGenie
12019
asked Jan 16 '13 at 3:26
EngGenieEngGenie
12019
12019
1
$begingroup$
Hint: $x < e^x$ so $I < ldots$.
$endgroup$
– Robert Israel
Jan 16 '13 at 4:12
add a comment |
1
$begingroup$
Hint: $x < e^x$ so $I < ldots$.
$endgroup$
– Robert Israel
Jan 16 '13 at 4:12
1
1
$begingroup$
Hint: $x < e^x$ so $I < ldots$.
$endgroup$
– Robert Israel
Jan 16 '13 at 4:12
$begingroup$
Hint: $x < e^x$ so $I < ldots$.
$endgroup$
– Robert Israel
Jan 16 '13 at 4:12
add a comment |
1 Answer
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$begingroup$
Robert's hint should answer your question.
I might go about the original problem a different way. Of $J$, $K$, and $L$, there is only one where the antiderivative is easy to compute, and it's easy to see whether that one converges. That observation eliminates several of the 6 alternatives given. Of those remaining, only one is not nonsense (e.g., (f) is nonsense since it deduces convergence from being bigger than something that converges).
answered Jan 16 '13 at 5:24
Gerry MyersonGerry Myerson
147k8147300
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1 Answer
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$begingroup$
Robert's hint should answer your question.
I might go about the original problem a different way. Of $J$, $K$, and $L$, there is only one where the antiderivative is easy to compute, and it's easy to see whether that one converges. That observation eliminates several of the 6 alternatives given. Of those remaining, only one is not nonsense (e.g., (f) is nonsense since it deduces convergence from being bigger than something that converges).
answered Jan 16 '13 at 5:24
Gerry MyersonGerry Myerson
147k8147300
$endgroup$
add a comment |
$begingroup$
Robert's hint should answer your question.
I might go about the original problem a different way. Of $J$, $K$, and $L$, there is only one where the antiderivative is easy to compute, and it's easy to see whether that one converges. That observation eliminates several of the 6 alternatives given. Of those remaining, only one is not nonsense (e.g., (f) is nonsense since it deduces convergence from being bigger than something that converges).
answered Jan 16 '13 at 5:24
Gerry MyersonGerry Myerson
147k8147300
$endgroup$
add a comment |
$begingroup$
Robert's hint should answer your question.
I might go about the original problem a different way. Of $J$, $K$, and $L$, there is only one where the antiderivative is easy to compute, and it's easy to see whether that one converges. That observation eliminates several of the 6 alternatives given. Of those remaining, only one is not nonsense (e.g., (f) is nonsense since it deduces convergence from being bigger than something that converges).
answered Jan 16 '13 at 5:24
Gerry MyersonGerry Myerson
147k8147300
$endgroup$
Robert's hint should answer your question.
I might go about the original problem a different way. Of $J$, $K$, and $L$, there is only one where the antiderivative is easy to compute, and it's easy to see whether that one converges. That observation eliminates several of the 6 alternatives given. Of those remaining, only one is not nonsense (e.g., (f) is nonsense since it deduces convergence from being bigger than something that converges).
answered Jan 16 '13 at 5:24
Gerry MyersonGerry Myerson
147k8147300
answered Jan 16 '13 at 5:24
Gerry MyersonGerry Myerson
147k8147300
answered Jan 16 '13 at 5:24
Gerry MyersonGerry Myerson
147k8147300
answered Jan 16 '13 at 5:24
Gerry MyersonGerry Myerson
147k8147300
147k8147300
add a comment |
add a comment |
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$begingroup$
Hint: $x < e^x$ so $I < ldots$.
$endgroup$
– Robert Israel
Jan 16 '13 at 4:12