Absolute integrability of $f(x) = frac{1+e^{|x|}}{1+x^2}$ over $mathbb{R}$
$begingroup$
I already have derived so far that $|f|=f$, and that $f$ is even. Thus, the integral we're ultimately concerned with is
$$2int_0^infty frac{1+e^x}{1+x^2}dx$$
My current instinct is that this integral is infinite, i.e. $f$ is not absolutely integrable. I just need to prove as much. This is not really a simple integral to evaluate, so I'm a little stumped.
My current approach is through the comparison test. Suppose we have a second function $g$ such that $| g(x) | leq | f(x) |$. Then if $g$ is not absolutely integrable, neither is $f$.
My goal is to try to get a second function $g$ that satisfies this. I first split up the integral into two integrals, and focus on the one with the exponential in it. (I've already calculated the former to be $pi/2$.)
$$2int_0^infty frac{1+e^x}{1+x^2}dx = 2left( int_0^infty frac{1}{1+x^2}dx +int_0^infty frac{e^x}{1+x^2}dx right ) = pi +int_0^infty frac{e^x}{1+x^2}dx $$
I've tried a number of functions: I want a function less than $frac{e^x}{1+x^2}$ that is also not a finite value to make this work. I've got heuristic arguments as for why $f$ should not be absolutely integrable, but I want be more formal.
Any ideas for a theorem I might be overlooking, or for a suitable function that actually has a nice antiderivative that would make this work?
(Sorry for all the edits you might have seen as I made this post. I originally deleted this when I saw that my entire approach in the previous time this was posted was ultimately incorrect owing to a flaw of my own.)
fourier-analysis improper-integrals
asked Oct 22 '18 at 22:49
Eevee TrainerEevee Trainer
6,67311237
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add a comment |
$begingroup$
I already have derived so far that $|f|=f$, and that $f$ is even. Thus, the integral we're ultimately concerned with is
$$2int_0^infty frac{1+e^x}{1+x^2}dx$$
My current instinct is that this integral is infinite, i.e. $f$ is not absolutely integrable. I just need to prove as much. This is not really a simple integral to evaluate, so I'm a little stumped.
My current approach is through the comparison test. Suppose we have a second function $g$ such that $| g(x) | leq | f(x) |$. Then if $g$ is not absolutely integrable, neither is $f$.
My goal is to try to get a second function $g$ that satisfies this. I first split up the integral into two integrals, and focus on the one with the exponential in it. (I've already calculated the former to be $pi/2$.)
$$2int_0^infty frac{1+e^x}{1+x^2}dx = 2left( int_0^infty frac{1}{1+x^2}dx +int_0^infty frac{e^x}{1+x^2}dx right ) = pi +int_0^infty frac{e^x}{1+x^2}dx $$
I've tried a number of functions: I want a function less than $frac{e^x}{1+x^2}$ that is also not a finite value to make this work. I've got heuristic arguments as for why $f$ should not be absolutely integrable, but I want be more formal.
Any ideas for a theorem I might be overlooking, or for a suitable function that actually has a nice antiderivative that would make this work?
(Sorry for all the edits you might have seen as I made this post. I originally deleted this when I saw that my entire approach in the previous time this was posted was ultimately incorrect owing to a flaw of my own.)
fourier-analysis improper-integrals
asked Oct 22 '18 at 22:49
Eevee TrainerEevee Trainer
6,67311237
$endgroup$
2
$begingroup$
$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$ diverges since $frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12$ (closer analysis shows the minimum is $frac e2$).
$endgroup$
– robjohn♦
Oct 22 '18 at 23:43
$begingroup$
I believe this follows from the power series definition of $e^x$, or does that follow from a separate identity?
$endgroup$
– Eevee Trainer
Oct 23 '18 at 0:00
$begingroup$
should it be $$pi+2int_0^inftyfrac{e^x}{1+x^2}dx$$
$endgroup$
– Henry Lee
Oct 23 '18 at 0:28
$begingroup$
The inequality in my comment follows from the first 3 terms of the power series, knowing that the rest of the terms are positive.
$endgroup$
– robjohn♦
Oct 23 '18 at 1:57
$begingroup$
@HenryLee Yeah, that's my bad, I got a bit hasty in copying everything. And thanks robjohn - I figured that was it but I wanted to double check.
$endgroup$
– Eevee Trainer
Oct 23 '18 at 4:13
add a comment |
$begingroup$
I already have derived so far that $|f|=f$, and that $f$ is even. Thus, the integral we're ultimately concerned with is
$$2int_0^infty frac{1+e^x}{1+x^2}dx$$
My current instinct is that this integral is infinite, i.e. $f$ is not absolutely integrable. I just need to prove as much. This is not really a simple integral to evaluate, so I'm a little stumped.
My current approach is through the comparison test. Suppose we have a second function $g$ such that $| g(x) | leq | f(x) |$. Then if $g$ is not absolutely integrable, neither is $f$.
My goal is to try to get a second function $g$ that satisfies this. I first split up the integral into two integrals, and focus on the one with the exponential in it. (I've already calculated the former to be $pi/2$.)
$$2int_0^infty frac{1+e^x}{1+x^2}dx = 2left( int_0^infty frac{1}{1+x^2}dx +int_0^infty frac{e^x}{1+x^2}dx right ) = pi +int_0^infty frac{e^x}{1+x^2}dx $$
I've tried a number of functions: I want a function less than $frac{e^x}{1+x^2}$ that is also not a finite value to make this work. I've got heuristic arguments as for why $f$ should not be absolutely integrable, but I want be more formal.
Any ideas for a theorem I might be overlooking, or for a suitable function that actually has a nice antiderivative that would make this work?
(Sorry for all the edits you might have seen as I made this post. I originally deleted this when I saw that my entire approach in the previous time this was posted was ultimately incorrect owing to a flaw of my own.)
fourier-analysis improper-integrals
asked Oct 22 '18 at 22:49
Eevee TrainerEevee Trainer
6,67311237
$endgroup$
I already have derived so far that $|f|=f$, and that $f$ is even. Thus, the integral we're ultimately concerned with is
$$2int_0^infty frac{1+e^x}{1+x^2}dx$$
My current instinct is that this integral is infinite, i.e. $f$ is not absolutely integrable. I just need to prove as much. This is not really a simple integral to evaluate, so I'm a little stumped.
My current approach is through the comparison test. Suppose we have a second function $g$ such that $| g(x) | leq | f(x) |$. Then if $g$ is not absolutely integrable, neither is $f$.
My goal is to try to get a second function $g$ that satisfies this. I first split up the integral into two integrals, and focus on the one with the exponential in it. (I've already calculated the former to be $pi/2$.)
$$2int_0^infty frac{1+e^x}{1+x^2}dx = 2left( int_0^infty frac{1}{1+x^2}dx +int_0^infty frac{e^x}{1+x^2}dx right ) = pi +int_0^infty frac{e^x}{1+x^2}dx $$
I've tried a number of functions: I want a function less than $frac{e^x}{1+x^2}$ that is also not a finite value to make this work. I've got heuristic arguments as for why $f$ should not be absolutely integrable, but I want be more formal.
Any ideas for a theorem I might be overlooking, or for a suitable function that actually has a nice antiderivative that would make this work?
(Sorry for all the edits you might have seen as I made this post. I originally deleted this when I saw that my entire approach in the previous time this was posted was ultimately incorrect owing to a flaw of my own.)
fourier-analysis improper-integrals
fourier-analysis improper-integrals
asked Oct 22 '18 at 22:49
Eevee TrainerEevee Trainer
6,67311237
asked Oct 22 '18 at 22:49
Eevee TrainerEevee Trainer
6,67311237
edited Oct 22 '18 at 23:28
Eevee Trainer
asked Oct 22 '18 at 22:49
Eevee TrainerEevee Trainer
6,67311237
asked Oct 22 '18 at 22:49
Eevee TrainerEevee Trainer
6,67311237
asked Oct 22 '18 at 22:49
Eevee TrainerEevee Trainer
6,67311237
6,67311237
2
$begingroup$
$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$ diverges since $frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12$ (closer analysis shows the minimum is $frac e2$).
$endgroup$
– robjohn♦
Oct 22 '18 at 23:43
$begingroup$
I believe this follows from the power series definition of $e^x$, or does that follow from a separate identity?
$endgroup$
– Eevee Trainer
Oct 23 '18 at 0:00
$begingroup$
should it be $$pi+2int_0^inftyfrac{e^x}{1+x^2}dx$$
$endgroup$
– Henry Lee
Oct 23 '18 at 0:28
$begingroup$
The inequality in my comment follows from the first 3 terms of the power series, knowing that the rest of the terms are positive.
$endgroup$
– robjohn♦
Oct 23 '18 at 1:57
$begingroup$
@HenryLee Yeah, that's my bad, I got a bit hasty in copying everything. And thanks robjohn - I figured that was it but I wanted to double check.
$endgroup$
– Eevee Trainer
Oct 23 '18 at 4:13
add a comment |
2
$begingroup$
$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$ diverges since $frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12$ (closer analysis shows the minimum is $frac e2$).
$endgroup$
– robjohn♦
Oct 22 '18 at 23:43
$begingroup$
I believe this follows from the power series definition of $e^x$, or does that follow from a separate identity?
$endgroup$
– Eevee Trainer
Oct 23 '18 at 0:00
$begingroup$
should it be $$pi+2int_0^inftyfrac{e^x}{1+x^2}dx$$
$endgroup$
– Henry Lee
Oct 23 '18 at 0:28
$begingroup$
The inequality in my comment follows from the first 3 terms of the power series, knowing that the rest of the terms are positive.
$endgroup$
– robjohn♦
Oct 23 '18 at 1:57
$begingroup$
@HenryLee Yeah, that's my bad, I got a bit hasty in copying everything. And thanks robjohn - I figured that was it but I wanted to double check.
$endgroup$
– Eevee Trainer
Oct 23 '18 at 4:13
2
2
$begingroup$
$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$ diverges since $frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12$ (closer analysis shows the minimum is $frac e2$).
$endgroup$
– robjohn♦
Oct 22 '18 at 23:43
$begingroup$
$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$ diverges since $frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12$ (closer analysis shows the minimum is $frac e2$).
$endgroup$
– robjohn♦
Oct 22 '18 at 23:43
$begingroup$
I believe this follows from the power series definition of $e^x$, or does that follow from a separate identity?
$endgroup$
– Eevee Trainer
Oct 23 '18 at 0:00
$begingroup$
I believe this follows from the power series definition of $e^x$, or does that follow from a separate identity?
$endgroup$
– Eevee Trainer
Oct 23 '18 at 0:00
$begingroup$
should it be $$pi+2int_0^inftyfrac{e^x}{1+x^2}dx$$
$endgroup$
– Henry Lee
Oct 23 '18 at 0:28
$begingroup$
should it be $$pi+2int_0^inftyfrac{e^x}{1+x^2}dx$$
$endgroup$
– Henry Lee
Oct 23 '18 at 0:28
$begingroup$
The inequality in my comment follows from the first 3 terms of the power series, knowing that the rest of the terms are positive.
$endgroup$
– robjohn♦
Oct 23 '18 at 1:57
$begingroup$
The inequality in my comment follows from the first 3 terms of the power series, knowing that the rest of the terms are positive.
$endgroup$
– robjohn♦
Oct 23 '18 at 1:57
$begingroup$
@HenryLee Yeah, that's my bad, I got a bit hasty in copying everything. And thanks robjohn - I figured that was it but I wanted to double check.
$endgroup$
– Eevee Trainer
Oct 23 '18 at 4:13
$begingroup$
@HenryLee Yeah, that's my bad, I got a bit hasty in copying everything. And thanks robjohn - I figured that was it but I wanted to double check.
$endgroup$
– Eevee Trainer
Oct 23 '18 at 4:13
add a comment |
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$begingroup$
Disclosure: I'm revisiting questions I asked during my past semester and closing some of those that never got an answer, at least an answer I can click "accept" on to close the question. In light of that, in this question, I'm going to (essentially) copy and paste the solution by robjohn from the comments of my question, which yielded the right solution, and let it be marked as a community answer.
Consider:
$$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$$
This integral diverges, because, following from the power series expansion of $e^x$,
$$ frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12 $$
$endgroup$
add a comment |
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$begingroup$
Disclosure: I'm revisiting questions I asked during my past semester and closing some of those that never got an answer, at least an answer I can click "accept" on to close the question. In light of that, in this question, I'm going to (essentially) copy and paste the solution by robjohn from the comments of my question, which yielded the right solution, and let it be marked as a community answer.
Consider:
$$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$$
This integral diverges, because, following from the power series expansion of $e^x$,
$$ frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12 $$
$endgroup$
add a comment |
$begingroup$
Disclosure: I'm revisiting questions I asked during my past semester and closing some of those that never got an answer, at least an answer I can click "accept" on to close the question. In light of that, in this question, I'm going to (essentially) copy and paste the solution by robjohn from the comments of my question, which yielded the right solution, and let it be marked as a community answer.
Consider:
$$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$$
This integral diverges, because, following from the power series expansion of $e^x$,
$$ frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12 $$
$endgroup$
add a comment |
$begingroup$
Disclosure: I'm revisiting questions I asked during my past semester and closing some of those that never got an answer, at least an answer I can click "accept" on to close the question. In light of that, in this question, I'm going to (essentially) copy and paste the solution by robjohn from the comments of my question, which yielded the right solution, and let it be marked as a community answer.
Consider:
$$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$$
This integral diverges, because, following from the power series expansion of $e^x$,
$$ frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12 $$
$endgroup$
Disclosure: I'm revisiting questions I asked during my past semester and closing some of those that never got an answer, at least an answer I can click "accept" on to close the question. In light of that, in this question, I'm going to (essentially) copy and paste the solution by robjohn from the comments of my question, which yielded the right solution, and let it be marked as a community answer.
Consider:
$$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$$
This integral diverges, because, following from the power series expansion of $e^x$,
$$ frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12 $$
answered Dec 30 '18 at 8:39
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Eevee Trainer
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$begingroup$
$int_0^inftyfrac{e^x}{1+x^2},mathrm{d}x$ diverges since $frac{e^x}{1+x^2}gefrac{1+x+frac12x^2}{1+x^2}gefrac12$ (closer analysis shows the minimum is $frac e2$).
$endgroup$
– robjohn♦
Oct 22 '18 at 23:43
$begingroup$
I believe this follows from the power series definition of $e^x$, or does that follow from a separate identity?
$endgroup$
– Eevee Trainer
Oct 23 '18 at 0:00
$begingroup$
should it be $$pi+2int_0^inftyfrac{e^x}{1+x^2}dx$$
$endgroup$
– Henry Lee
Oct 23 '18 at 0:28
$begingroup$
The inequality in my comment follows from the first 3 terms of the power series, knowing that the rest of the terms are positive.
$endgroup$
– robjohn♦
Oct 23 '18 at 1:57
$begingroup$
@HenryLee Yeah, that's my bad, I got a bit hasty in copying everything. And thanks robjohn - I figured that was it but I wanted to double check.
$endgroup$
– Eevee Trainer
Oct 23 '18 at 4:13