Proving that $int_{-infty}^{infty}frac{cos(x)}{x^2}dx$ is convergent/divergent?
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I want to determine whether $int_{-infty}^{infty}frac{cos(x)}{x^2}dx$ is convergent or not. I first said that, for any $x$, $frac{cos(x)}{x^2} ge frac{-10}{x^2}$. Since $frac{-10}{x^2}$ is divergent, then the initial integral is too. However, the integral calculator website says that $int_{0}^{infty}frac{cos(x)}{x^2}dx$ and $int_{-infty}^{0}frac{cos(x)}{x^2}dx$ are both convergent. Wouldn't it then follow that the initial integral is also convergent?
What error am I making? Thank you.
integration
asked Dec 5 at 2:02
user494405
37219
add a comment |
up vote
0
down vote
favorite
I want to determine whether $int_{-infty}^{infty}frac{cos(x)}{x^2}dx$ is convergent or not. I first said that, for any $x$, $frac{cos(x)}{x^2} ge frac{-10}{x^2}$. Since $frac{-10}{x^2}$ is divergent, then the initial integral is too. However, the integral calculator website says that $int_{0}^{infty}frac{cos(x)}{x^2}dx$ and $int_{-infty}^{0}frac{cos(x)}{x^2}dx$ are both convergent. Wouldn't it then follow that the initial integral is also convergent?
What error am I making? Thank you.
integration
asked Dec 5 at 2:02
user494405
37219
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to determine whether $int_{-infty}^{infty}frac{cos(x)}{x^2}dx$ is convergent or not. I first said that, for any $x$, $frac{cos(x)}{x^2} ge frac{-10}{x^2}$. Since $frac{-10}{x^2}$ is divergent, then the initial integral is too. However, the integral calculator website says that $int_{0}^{infty}frac{cos(x)}{x^2}dx$ and $int_{-infty}^{0}frac{cos(x)}{x^2}dx$ are both convergent. Wouldn't it then follow that the initial integral is also convergent?
What error am I making? Thank you.
integration
asked Dec 5 at 2:02
user494405
37219
I want to determine whether $int_{-infty}^{infty}frac{cos(x)}{x^2}dx$ is convergent or not. I first said that, for any $x$, $frac{cos(x)}{x^2} ge frac{-10}{x^2}$. Since $frac{-10}{x^2}$ is divergent, then the initial integral is too. However, the integral calculator website says that $int_{0}^{infty}frac{cos(x)}{x^2}dx$ and $int_{-infty}^{0}frac{cos(x)}{x^2}dx$ are both convergent. Wouldn't it then follow that the initial integral is also convergent?
What error am I making? Thank you.
integration
integration
asked Dec 5 at 2:02
user494405
37219
asked Dec 5 at 2:02
user494405
37219
edited Dec 5 at 2:16
asked Dec 5 at 2:02
user494405
37219
asked Dec 5 at 2:02
user494405
37219
asked Dec 5 at 2:02
user494405
37219
37219
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
That
$$
frac{cos x}{x^2}leq frac{-10}{x^2}
$$
for any $x$ is certainly not true, take $x=2pi$ among many other choices.
Also, bounding something above by a divergent integral tells you nothing. Also, depending on where you are examining $1/x^2$, it can be integrable, like at infinity.
However, you were right by accident. It does not converge, and your website is wrong.
$$
cos(x)=1+O(x^2)
$$
so,
$$
int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}
$$
which is not finite. So, there is no hope for $int_0^infty frac{cos x}{x^2}mathrm dx$ or $int_{-infty}^0 frac{cos x}{x^2}mathrm dx$.
answered Dec 5 at 2:14
qbert
22k32459
My mistake, I actually meant to write $frac{cos x}{x^2}geq frac{-10}{x^2}$, hence why I chose a divergent integral. I've fixed my post. With that in mind, does my logic work? Thanks.
– user494405
Dec 5 at 2:18
1
Still no, you have bounded it from below by $-infty$. You want to push it off to infinity using something that diverges to positive infinity
– qbert
Dec 5 at 2:20
Ah, of course, how dumb of me. So it's enough to say that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$, because, if an integral is positively divergent within a certain interval, then it is of course divergent on any superset of that interval. Correct? Thanks again.
– user494405
Dec 5 at 2:25
1
@user494405 that's right. Not a problem
– qbert
Dec 5 at 2:25
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
– user494405
Dec 5 at 2:34
|
show 1 more comment
up vote
1
down vote
You might want to note that integral calculator websites can be inaccurate. I'm not saying it isn't, here, but depending on the way in the integral converges or diverges, the integral can be handled weirdly by the calculator.
Also, I will take some issue with a bit of your logic: generally speaking, for a pair of functions $f,g$ on an interval $(alpha,beta)$, if $|f(x)| leq |g(x)|$, then we can say two things:
If $int_alpha^beta |g(x)|dx$ is a finite real value (what we call $g$ being "absolutely integrable" on that interval), then so is $f$
If $int_alpha^beta |f(x)|dx$ diverges, the same must be true for $g$.
Your example with $frac{cos(x)}{x^2} leq -frac{10}{x^2}$ and the latter diverging doesn't mean anything in this case. (The inequality is also not true. As you can see, it implies $cos(x) leq -10$, but $cos(x)in [-1,1] ; forall x in mathbb{R}$.)
I'm not 100% sure how you should calculate it. My personal guess would utilizing the power series for cosine but it's just that, a guess.
answered Dec 5 at 2:15
Eevee Trainer
3,295224
add a comment |
up vote
0
down vote
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
but $C=0$ so:
$$I(a)=pi|a|$$
which would give $I=pi$, whilst I believe the correct answer is $I=0$ so this must be wrong somewhere
answered Dec 5 at 2:40
Henry Lee
1,707218
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
That
$$
frac{cos x}{x^2}leq frac{-10}{x^2}
$$
for any $x$ is certainly not true, take $x=2pi$ among many other choices.
Also, bounding something above by a divergent integral tells you nothing. Also, depending on where you are examining $1/x^2$, it can be integrable, like at infinity.
However, you were right by accident. It does not converge, and your website is wrong.
$$
cos(x)=1+O(x^2)
$$
so,
$$
int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}
$$
which is not finite. So, there is no hope for $int_0^infty frac{cos x}{x^2}mathrm dx$ or $int_{-infty}^0 frac{cos x}{x^2}mathrm dx$.
answered Dec 5 at 2:14
qbert
22k32459
My mistake, I actually meant to write $frac{cos x}{x^2}geq frac{-10}{x^2}$, hence why I chose a divergent integral. I've fixed my post. With that in mind, does my logic work? Thanks.
– user494405
Dec 5 at 2:18
1
Still no, you have bounded it from below by $-infty$. You want to push it off to infinity using something that diverges to positive infinity
– qbert
Dec 5 at 2:20
Ah, of course, how dumb of me. So it's enough to say that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$, because, if an integral is positively divergent within a certain interval, then it is of course divergent on any superset of that interval. Correct? Thanks again.
– user494405
Dec 5 at 2:25
1
@user494405 that's right. Not a problem
– qbert
Dec 5 at 2:25
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
– user494405
Dec 5 at 2:34
|
show 1 more comment
up vote
2
down vote
accepted
That
$$
frac{cos x}{x^2}leq frac{-10}{x^2}
$$
for any $x$ is certainly not true, take $x=2pi$ among many other choices.
Also, bounding something above by a divergent integral tells you nothing. Also, depending on where you are examining $1/x^2$, it can be integrable, like at infinity.
However, you were right by accident. It does not converge, and your website is wrong.
$$
cos(x)=1+O(x^2)
$$
so,
$$
int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}
$$
which is not finite. So, there is no hope for $int_0^infty frac{cos x}{x^2}mathrm dx$ or $int_{-infty}^0 frac{cos x}{x^2}mathrm dx$.
answered Dec 5 at 2:14
qbert
22k32459
My mistake, I actually meant to write $frac{cos x}{x^2}geq frac{-10}{x^2}$, hence why I chose a divergent integral. I've fixed my post. With that in mind, does my logic work? Thanks.
– user494405
Dec 5 at 2:18
1
Still no, you have bounded it from below by $-infty$. You want to push it off to infinity using something that diverges to positive infinity
– qbert
Dec 5 at 2:20
Ah, of course, how dumb of me. So it's enough to say that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$, because, if an integral is positively divergent within a certain interval, then it is of course divergent on any superset of that interval. Correct? Thanks again.
– user494405
Dec 5 at 2:25
1
@user494405 that's right. Not a problem
– qbert
Dec 5 at 2:25
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
– user494405
Dec 5 at 2:34
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
That
$$
frac{cos x}{x^2}leq frac{-10}{x^2}
$$
for any $x$ is certainly not true, take $x=2pi$ among many other choices.
Also, bounding something above by a divergent integral tells you nothing. Also, depending on where you are examining $1/x^2$, it can be integrable, like at infinity.
However, you were right by accident. It does not converge, and your website is wrong.
$$
cos(x)=1+O(x^2)
$$
so,
$$
int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}
$$
which is not finite. So, there is no hope for $int_0^infty frac{cos x}{x^2}mathrm dx$ or $int_{-infty}^0 frac{cos x}{x^2}mathrm dx$.
answered Dec 5 at 2:14
qbert
22k32459
That
$$
frac{cos x}{x^2}leq frac{-10}{x^2}
$$
for any $x$ is certainly not true, take $x=2pi$ among many other choices.
Also, bounding something above by a divergent integral tells you nothing. Also, depending on where you are examining $1/x^2$, it can be integrable, like at infinity.
However, you were right by accident. It does not converge, and your website is wrong.
$$
cos(x)=1+O(x^2)
$$
so,
$$
int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}
$$
which is not finite. So, there is no hope for $int_0^infty frac{cos x}{x^2}mathrm dx$ or $int_{-infty}^0 frac{cos x}{x^2}mathrm dx$.
answered Dec 5 at 2:14
qbert
22k32459
edited Dec 5 at 2:19
answered Dec 5 at 2:14
qbert
22k32459
answered Dec 5 at 2:14
qbert
22k32459
answered Dec 5 at 2:14
qbert
22k32459
22k32459
My mistake, I actually meant to write $frac{cos x}{x^2}geq frac{-10}{x^2}$, hence why I chose a divergent integral. I've fixed my post. With that in mind, does my logic work? Thanks.
– user494405
Dec 5 at 2:18
1
Still no, you have bounded it from below by $-infty$. You want to push it off to infinity using something that diverges to positive infinity
– qbert
Dec 5 at 2:20
Ah, of course, how dumb of me. So it's enough to say that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$, because, if an integral is positively divergent within a certain interval, then it is of course divergent on any superset of that interval. Correct? Thanks again.
– user494405
Dec 5 at 2:25
1
@user494405 that's right. Not a problem
– qbert
Dec 5 at 2:25
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
– user494405
Dec 5 at 2:34
|
show 1 more comment
My mistake, I actually meant to write $frac{cos x}{x^2}geq frac{-10}{x^2}$, hence why I chose a divergent integral. I've fixed my post. With that in mind, does my logic work? Thanks.
– user494405
Dec 5 at 2:18
1
Still no, you have bounded it from below by $-infty$. You want to push it off to infinity using something that diverges to positive infinity
– qbert
Dec 5 at 2:20
Ah, of course, how dumb of me. So it's enough to say that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$, because, if an integral is positively divergent within a certain interval, then it is of course divergent on any superset of that interval. Correct? Thanks again.
– user494405
Dec 5 at 2:25
1
@user494405 that's right. Not a problem
– qbert
Dec 5 at 2:25
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
– user494405
Dec 5 at 2:34
My mistake, I actually meant to write $frac{cos x}{x^2}geq frac{-10}{x^2}$, hence why I chose a divergent integral. I've fixed my post. With that in mind, does my logic work? Thanks.
– user494405
Dec 5 at 2:18
My mistake, I actually meant to write $frac{cos x}{x^2}geq frac{-10}{x^2}$, hence why I chose a divergent integral. I've fixed my post. With that in mind, does my logic work? Thanks.
– user494405
Dec 5 at 2:18
1
1
Still no, you have bounded it from below by $-infty$. You want to push it off to infinity using something that diverges to positive infinity
– qbert
Dec 5 at 2:20
Still no, you have bounded it from below by $-infty$. You want to push it off to infinity using something that diverges to positive infinity
– qbert
Dec 5 at 2:20
Ah, of course, how dumb of me. So it's enough to say that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$, because, if an integral is positively divergent within a certain interval, then it is of course divergent on any superset of that interval. Correct? Thanks again.
– user494405
Dec 5 at 2:25
Ah, of course, how dumb of me. So it's enough to say that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$, because, if an integral is positively divergent within a certain interval, then it is of course divergent on any superset of that interval. Correct? Thanks again.
– user494405
Dec 5 at 2:25
1
1
@user494405 that's right. Not a problem
– qbert
Dec 5 at 2:25
@user494405 that's right. Not a problem
– qbert
Dec 5 at 2:25
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
– user494405
Dec 5 at 2:34
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
– user494405
Dec 5 at 2:34
|
show 1 more comment
up vote
1
down vote
You might want to note that integral calculator websites can be inaccurate. I'm not saying it isn't, here, but depending on the way in the integral converges or diverges, the integral can be handled weirdly by the calculator.
Also, I will take some issue with a bit of your logic: generally speaking, for a pair of functions $f,g$ on an interval $(alpha,beta)$, if $|f(x)| leq |g(x)|$, then we can say two things:
If $int_alpha^beta |g(x)|dx$ is a finite real value (what we call $g$ being "absolutely integrable" on that interval), then so is $f$
If $int_alpha^beta |f(x)|dx$ diverges, the same must be true for $g$.
Your example with $frac{cos(x)}{x^2} leq -frac{10}{x^2}$ and the latter diverging doesn't mean anything in this case. (The inequality is also not true. As you can see, it implies $cos(x) leq -10$, but $cos(x)in [-1,1] ; forall x in mathbb{R}$.)
I'm not 100% sure how you should calculate it. My personal guess would utilizing the power series for cosine but it's just that, a guess.
answered Dec 5 at 2:15
Eevee Trainer
3,295224
add a comment |
up vote
1
down vote
You might want to note that integral calculator websites can be inaccurate. I'm not saying it isn't, here, but depending on the way in the integral converges or diverges, the integral can be handled weirdly by the calculator.
Also, I will take some issue with a bit of your logic: generally speaking, for a pair of functions $f,g$ on an interval $(alpha,beta)$, if $|f(x)| leq |g(x)|$, then we can say two things:
If $int_alpha^beta |g(x)|dx$ is a finite real value (what we call $g$ being "absolutely integrable" on that interval), then so is $f$
If $int_alpha^beta |f(x)|dx$ diverges, the same must be true for $g$.
Your example with $frac{cos(x)}{x^2} leq -frac{10}{x^2}$ and the latter diverging doesn't mean anything in this case. (The inequality is also not true. As you can see, it implies $cos(x) leq -10$, but $cos(x)in [-1,1] ; forall x in mathbb{R}$.)
I'm not 100% sure how you should calculate it. My personal guess would utilizing the power series for cosine but it's just that, a guess.
answered Dec 5 at 2:15
Eevee Trainer
3,295224
add a comment |
up vote
1
down vote
up vote
1
down vote
You might want to note that integral calculator websites can be inaccurate. I'm not saying it isn't, here, but depending on the way in the integral converges or diverges, the integral can be handled weirdly by the calculator.
Also, I will take some issue with a bit of your logic: generally speaking, for a pair of functions $f,g$ on an interval $(alpha,beta)$, if $|f(x)| leq |g(x)|$, then we can say two things:
If $int_alpha^beta |g(x)|dx$ is a finite real value (what we call $g$ being "absolutely integrable" on that interval), then so is $f$
If $int_alpha^beta |f(x)|dx$ diverges, the same must be true for $g$.
Your example with $frac{cos(x)}{x^2} leq -frac{10}{x^2}$ and the latter diverging doesn't mean anything in this case. (The inequality is also not true. As you can see, it implies $cos(x) leq -10$, but $cos(x)in [-1,1] ; forall x in mathbb{R}$.)
I'm not 100% sure how you should calculate it. My personal guess would utilizing the power series for cosine but it's just that, a guess.
answered Dec 5 at 2:15
Eevee Trainer
3,295224
You might want to note that integral calculator websites can be inaccurate. I'm not saying it isn't, here, but depending on the way in the integral converges or diverges, the integral can be handled weirdly by the calculator.
Also, I will take some issue with a bit of your logic: generally speaking, for a pair of functions $f,g$ on an interval $(alpha,beta)$, if $|f(x)| leq |g(x)|$, then we can say two things:
If $int_alpha^beta |g(x)|dx$ is a finite real value (what we call $g$ being "absolutely integrable" on that interval), then so is $f$
If $int_alpha^beta |f(x)|dx$ diverges, the same must be true for $g$.
Your example with $frac{cos(x)}{x^2} leq -frac{10}{x^2}$ and the latter diverging doesn't mean anything in this case. (The inequality is also not true. As you can see, it implies $cos(x) leq -10$, but $cos(x)in [-1,1] ; forall x in mathbb{R}$.)
I'm not 100% sure how you should calculate it. My personal guess would utilizing the power series for cosine but it's just that, a guess.
answered Dec 5 at 2:15
Eevee Trainer
3,295224
answered Dec 5 at 2:15
Eevee Trainer
3,295224
answered Dec 5 at 2:15
Eevee Trainer
3,295224
answered Dec 5 at 2:15
Eevee Trainer
3,295224
3,295224
add a comment |
add a comment |
up vote
0
down vote
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
but $C=0$ so:
$$I(a)=pi|a|$$
which would give $I=pi$, whilst I believe the correct answer is $I=0$ so this must be wrong somewhere
answered Dec 5 at 2:40
Henry Lee
1,707218
add a comment |
up vote
0
down vote
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
but $C=0$ so:
$$I(a)=pi|a|$$
which would give $I=pi$, whilst I believe the correct answer is $I=0$ so this must be wrong somewhere
answered Dec 5 at 2:40
Henry Lee
1,707218
add a comment |
up vote
0
down vote
up vote
0
down vote
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
but $C=0$ so:
$$I(a)=pi|a|$$
which would give $I=pi$, whilst I believe the correct answer is $I=0$ so this must be wrong somewhere
answered Dec 5 at 2:40
Henry Lee
1,707218
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
but $C=0$ so:
$$I(a)=pi|a|$$
which would give $I=pi$, whilst I believe the correct answer is $I=0$ so this must be wrong somewhere
answered Dec 5 at 2:40
Henry Lee
1,707218
answered Dec 5 at 2:40
Henry Lee
1,707218
answered Dec 5 at 2:40
Henry Lee
1,707218
answered Dec 5 at 2:40
Henry Lee
1,707218
1,707218
add a comment |
add a comment |
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