Proving that $int_{-infty}^{infty}frac{cos(x)}{x^2}dx$ is convergent/divergent?
$begingroup$
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 '18 at 2:02
user494405user494405
37719
$endgroup$
add a comment |
$begingroup$
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 '18 at 2:02
user494405user494405
37719
$endgroup$
add a comment |
$begingroup$
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 '18 at 2:02
user494405user494405
37719
$endgroup$
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 '18 at 2:02
user494405user494405
37719
asked Dec 5 '18 at 2:02
user494405user494405
37719
edited Dec 5 '18 at 2:16
user494405
asked Dec 5 '18 at 2:02
user494405user494405
37719
asked Dec 5 '18 at 2:02
user494405user494405
37719
asked Dec 5 '18 at 2:02
user494405user494405
37719
37719
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 '18 at 2:14
qbertqbert
22.1k32560
$endgroup$
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:18
1
$begingroup$
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
$endgroup$
– qbert
Dec 5 '18 at 2:20
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:25
1
$begingroup$
@user494405 that's right. Not a problem
$endgroup$
– qbert
Dec 5 '18 at 2:25
$begingroup$
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
$endgroup$
– user494405
Dec 5 '18 at 2:34
|
show 1 more comment
$begingroup$
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 '18 at 2:15
Eevee TrainerEevee Trainer
5,8011936
$endgroup$
add a comment |
$begingroup$
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
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 '18 at 2:40
Henry LeeHenry Lee
1,884219
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
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active
oldest
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active
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active
oldest
votes
$begingroup$
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 '18 at 2:14
qbertqbert
22.1k32560
$endgroup$
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:18
1
$begingroup$
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
$endgroup$
– qbert
Dec 5 '18 at 2:20
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:25
1
$begingroup$
@user494405 that's right. Not a problem
$endgroup$
– qbert
Dec 5 '18 at 2:25
$begingroup$
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
$endgroup$
– user494405
Dec 5 '18 at 2:34
|
show 1 more comment
$begingroup$
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 '18 at 2:14
qbertqbert
22.1k32560
$endgroup$
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:18
1
$begingroup$
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
$endgroup$
– qbert
Dec 5 '18 at 2:20
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:25
1
$begingroup$
@user494405 that's right. Not a problem
$endgroup$
– qbert
Dec 5 '18 at 2:25
$begingroup$
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
$endgroup$
– user494405
Dec 5 '18 at 2:34
|
show 1 more comment
$begingroup$
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 '18 at 2:14
qbertqbert
22.1k32560
$endgroup$
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 '18 at 2:14
qbertqbert
22.1k32560
edited Dec 5 '18 at 2:19
answered Dec 5 '18 at 2:14
qbertqbert
22.1k32560
answered Dec 5 '18 at 2:14
qbertqbert
22.1k32560
answered Dec 5 '18 at 2:14
qbertqbert
22.1k32560
22.1k32560
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:18
1
$begingroup$
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
$endgroup$
– qbert
Dec 5 '18 at 2:20
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:25
1
$begingroup$
@user494405 that's right. Not a problem
$endgroup$
– qbert
Dec 5 '18 at 2:25
$begingroup$
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
$endgroup$
– user494405
Dec 5 '18 at 2:34
|
show 1 more comment
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:18
1
$begingroup$
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
$endgroup$
– qbert
Dec 5 '18 at 2:20
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:25
1
$begingroup$
@user494405 that's right. Not a problem
$endgroup$
– qbert
Dec 5 '18 at 2:25
$begingroup$
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
$endgroup$
– user494405
Dec 5 '18 at 2:34
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:18
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:18
1
1
$begingroup$
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
$endgroup$
– qbert
Dec 5 '18 at 2:20
$begingroup$
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
$endgroup$
– qbert
Dec 5 '18 at 2:20
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:25
$begingroup$
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.
$endgroup$
– user494405
Dec 5 '18 at 2:25
1
1
$begingroup$
@user494405 that's right. Not a problem
$endgroup$
– qbert
Dec 5 '18 at 2:25
$begingroup$
@user494405 that's right. Not a problem
$endgroup$
– qbert
Dec 5 '18 at 2:25
$begingroup$
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
$endgroup$
– user494405
Dec 5 '18 at 2:34
$begingroup$
One last question, if you don't mind. Why does $cos(x)=1+O(x^2)$ imply that $int_0^1frac{cos x}{x^2}mathrm dxapprox int_0^1 frac{1}{x^2}$?
$endgroup$
– user494405
Dec 5 '18 at 2:34
|
show 1 more comment
$begingroup$
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 '18 at 2:15
Eevee TrainerEevee Trainer
5,8011936
$endgroup$
add a comment |
$begingroup$
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 '18 at 2:15
Eevee TrainerEevee Trainer
5,8011936
$endgroup$
add a comment |
$begingroup$
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 '18 at 2:15
Eevee TrainerEevee Trainer
5,8011936
$endgroup$
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 '18 at 2:15
Eevee TrainerEevee Trainer
5,8011936
answered Dec 5 '18 at 2:15
Eevee TrainerEevee Trainer
5,8011936
answered Dec 5 '18 at 2:15
Eevee TrainerEevee Trainer
5,8011936
answered Dec 5 '18 at 2:15
Eevee TrainerEevee Trainer
5,8011936
5,8011936
add a comment |
add a comment |
$begingroup$
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
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 '18 at 2:40
Henry LeeHenry Lee
1,884219
$endgroup$
add a comment |
$begingroup$
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
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 '18 at 2:40
Henry LeeHenry Lee
1,884219
$endgroup$
add a comment |
$begingroup$
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
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 '18 at 2:40
Henry LeeHenry Lee
1,884219
$endgroup$
$$I(a)=int_{-infty}^inftyfrac{cos(ax)}{x^2}dx$$
$$I'(a)=int_{-infty}^inftyfrac{-sin(ax)}{x}dx=pitext{sgn}(a)$$
$$I(a)=intpitext{sgn}(a)da=pi|a|+C$$
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 '18 at 2:40
Henry LeeHenry Lee
1,884219
answered Dec 5 '18 at 2:40
Henry LeeHenry Lee
1,884219
answered Dec 5 '18 at 2:40
Henry LeeHenry Lee
1,884219
answered Dec 5 '18 at 2:40
Henry LeeHenry Lee
1,884219
1,884219
add a comment |
add a comment |
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