Mixing
How to match 2 integrals as a parameter goes to zero makes them the same integral [duplicate]
$begingroup$
This question already has an answer here:
Can a limit of an integral be moved inside the integral?
2 answers
I am interested in the integral $$int{dfrac{1}{r^3}}dx$$ with $ r=(x^2+a^2)^{1/2}$ That is $$int{dfrac{1}{(x^2+a^2)^{3/2}}}dx$$ A book shows $$int{dfrac{1}{(x^2+a^2)^{3/2}}}dx=dfrac{x}{a^2(x^2+a^2)^{1/2}}$$ In my case, $x$ will always be positive and real. The parameter $a$ is real and may be zero. In that case ($a=0$) I have $$int{dfrac{1}{(x^2)^{3/2}}}dx=dfrac{x}{0(x^2)^{1/2}}=dfrac{1}{0}$$ But this could also be written, with $a=0$, as $$int{dfrac{1}{x^3}}dx=-dfrac{1}{2x^2}$$ What am I missing that the two integrals give different answers for the case $a=0$? How can I make the two integrals agree? I need to be able to vary the parameter $a$ from zero to any positive real number. I need to understand how to handle the case that $a$ is very close to zero, as well as being zero exactly.
Thanks.
integration
asked Jan 6 at 16:30
user458534user458534
53
$endgroup$
marked as duplicate by amWhy integration
Users with the integration badge can single-handedly close integration questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 6 at 20:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Can a limit of an integral be moved inside the integral?
2 answers
I am interested in the integral $$int{dfrac{1}{r^3}}dx$$ with $ r=(x^2+a^2)^{1/2}$ That is $$int{dfrac{1}{(x^2+a^2)^{3/2}}}dx$$ A book shows $$int{dfrac{1}{(x^2+a^2)^{3/2}}}dx=dfrac{x}{a^2(x^2+a^2)^{1/2}}$$ In my case, $x$ will always be positive and real. The parameter $a$ is real and may be zero. In that case ($a=0$) I have $$int{dfrac{1}{(x^2)^{3/2}}}dx=dfrac{x}{0(x^2)^{1/2}}=dfrac{1}{0}$$ But this could also be written, with $a=0$, as $$int{dfrac{1}{x^3}}dx=-dfrac{1}{2x^2}$$ What am I missing that the two integrals give different answers for the case $a=0$? How can I make the two integrals agree? I need to be able to vary the parameter $a$ from zero to any positive real number. I need to understand how to handle the case that $a$ is very close to zero, as well as being zero exactly.
Thanks.
integration
asked Jan 6 at 16:30
user458534user458534
53
$endgroup$
marked as duplicate by amWhy integration
Users with the integration badge can single-handedly close integration questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 6 at 20:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
You are trying to move a limit inside an integral. Read this for enlightenment: math.stackexchange.com/q/253696/73324
$endgroup$
– vadim123
Jan 6 at 16:33
add a comment |
$begingroup$
This question already has an answer here:
Can a limit of an integral be moved inside the integral?
2 answers
I am interested in the integral $$int{dfrac{1}{r^3}}dx$$ with $ r=(x^2+a^2)^{1/2}$ That is $$int{dfrac{1}{(x^2+a^2)^{3/2}}}dx$$ A book shows $$int{dfrac{1}{(x^2+a^2)^{3/2}}}dx=dfrac{x}{a^2(x^2+a^2)^{1/2}}$$ In my case, $x$ will always be positive and real. The parameter $a$ is real and may be zero. In that case ($a=0$) I have $$int{dfrac{1}{(x^2)^{3/2}}}dx=dfrac{x}{0(x^2)^{1/2}}=dfrac{1}{0}$$ But this could also be written, with $a=0$, as $$int{dfrac{1}{x^3}}dx=-dfrac{1}{2x^2}$$ What am I missing that the two integrals give different answers for the case $a=0$? How can I make the two integrals agree? I need to be able to vary the parameter $a$ from zero to any positive real number. I need to understand how to handle the case that $a$ is very close to zero, as well as being zero exactly.
Thanks.
integration
asked Jan 6 at 16:30
user458534user458534
53
$endgroup$
This question already has an answer here:
Can a limit of an integral be moved inside the integral?
2 answers
I am interested in the integral $$int{dfrac{1}{r^3}}dx$$ with $ r=(x^2+a^2)^{1/2}$ That is $$int{dfrac{1}{(x^2+a^2)^{3/2}}}dx$$ A book shows $$int{dfrac{1}{(x^2+a^2)^{3/2}}}dx=dfrac{x}{a^2(x^2+a^2)^{1/2}}$$ In my case, $x$ will always be positive and real. The parameter $a$ is real and may be zero. In that case ($a=0$) I have $$int{dfrac{1}{(x^2)^{3/2}}}dx=dfrac{x}{0(x^2)^{1/2}}=dfrac{1}{0}$$ But this could also be written, with $a=0$, as $$int{dfrac{1}{x^3}}dx=-dfrac{1}{2x^2}$$ What am I missing that the two integrals give different answers for the case $a=0$? How can I make the two integrals agree? I need to be able to vary the parameter $a$ from zero to any positive real number. I need to understand how to handle the case that $a$ is very close to zero, as well as being zero exactly.
Thanks.
This question already has an answer here:
Can a limit of an integral be moved inside the integral?
2 answers
integration
integration
asked Jan 6 at 16:30
user458534user458534
53
asked Jan 6 at 16:30
user458534user458534
53
asked Jan 6 at 16:30
user458534user458534
53
asked Jan 6 at 16:30
user458534user458534
53
asked Jan 6 at 16:30
user458534user458534
53
53
marked as duplicate by amWhy integration
Users with the integration badge can single-handedly close integration questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 6 at 20:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by amWhy integration
Users with the integration badge can single-handedly close integration questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 6 at 20:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
You are trying to move a limit inside an integral. Read this for enlightenment: math.stackexchange.com/q/253696/73324
$endgroup$
– vadim123
Jan 6 at 16:33
add a comment |
1
$begingroup$
You are trying to move a limit inside an integral. Read this for enlightenment: math.stackexchange.com/q/253696/73324
$endgroup$
– vadim123
Jan 6 at 16:33
1
1
$begingroup$
You are trying to move a limit inside an integral. Read this for enlightenment: math.stackexchange.com/q/253696/73324
$endgroup$
– vadim123
Jan 6 at 16:33
$begingroup$
You are trying to move a limit inside an integral. Read this for enlightenment: math.stackexchange.com/q/253696/73324
$endgroup$
– vadim123
Jan 6 at 16:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:Substitute $$x=atan(u)$$ then $$dx=asec^2(u)du$$ and your integral is given by $$aintfrac{cos(u)}{a^3}du$$
The result is given by $$frac{x}{a^2 sqrt{a^2+x^2}}+C$$
and the limit for $a$ tends to zero doesn't exist.
answered Jan 6 at 16:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.4k42865
$endgroup$
$begingroup$
This does help to derive the integral, but this doesn't answer the further question of why the two answers are different in $lim_{ato0}$.
$endgroup$
– John Doe
Jan 6 at 16:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:Substitute $$x=atan(u)$$ then $$dx=asec^2(u)du$$ and your integral is given by $$aintfrac{cos(u)}{a^3}du$$
The result is given by $$frac{x}{a^2 sqrt{a^2+x^2}}+C$$
and the limit for $a$ tends to zero doesn't exist.
answered Jan 6 at 16:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.4k42865
$endgroup$
$begingroup$
This does help to derive the integral, but this doesn't answer the further question of why the two answers are different in $lim_{ato0}$.
$endgroup$
– John Doe
Jan 6 at 16:43
add a comment |
$begingroup$
Hint:Substitute $$x=atan(u)$$ then $$dx=asec^2(u)du$$ and your integral is given by $$aintfrac{cos(u)}{a^3}du$$
The result is given by $$frac{x}{a^2 sqrt{a^2+x^2}}+C$$
and the limit for $a$ tends to zero doesn't exist.
answered Jan 6 at 16:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.4k42865
$endgroup$
$begingroup$
This does help to derive the integral, but this doesn't answer the further question of why the two answers are different in $lim_{ato0}$.
$endgroup$
– John Doe
Jan 6 at 16:43
add a comment |
$begingroup$
Hint:Substitute $$x=atan(u)$$ then $$dx=asec^2(u)du$$ and your integral is given by $$aintfrac{cos(u)}{a^3}du$$
The result is given by $$frac{x}{a^2 sqrt{a^2+x^2}}+C$$
and the limit for $a$ tends to zero doesn't exist.
answered Jan 6 at 16:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.4k42865
$endgroup$
Hint:Substitute $$x=atan(u)$$ then $$dx=asec^2(u)du$$ and your integral is given by $$aintfrac{cos(u)}{a^3}du$$
The result is given by $$frac{x}{a^2 sqrt{a^2+x^2}}+C$$
and the limit for $a$ tends to zero doesn't exist.
answered Jan 6 at 16:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.4k42865
edited Jan 6 at 16:47
answered Jan 6 at 16:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.4k42865
answered Jan 6 at 16:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.4k42865
answered Jan 6 at 16:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.4k42865
74.4k42865
$begingroup$
This does help to derive the integral, but this doesn't answer the further question of why the two answers are different in $lim_{ato0}$.
$endgroup$
– John Doe
Jan 6 at 16:43
add a comment |
$begingroup$
This does help to derive the integral, but this doesn't answer the further question of why the two answers are different in $lim_{ato0}$.
$endgroup$
– John Doe
Jan 6 at 16:43
$begingroup$
This does help to derive the integral, but this doesn't answer the further question of why the two answers are different in $lim_{ato0}$.
$endgroup$
– John Doe
Jan 6 at 16:43
$begingroup$
This does help to derive the integral, but this doesn't answer the further question of why the two answers are different in $lim_{ato0}$.
$endgroup$
– John Doe
Jan 6 at 16:43
add a comment |
1
$begingroup$
You are trying to move a limit inside an integral. Read this for enlightenment: math.stackexchange.com/q/253696/73324
$endgroup$
– vadim123
Jan 6 at 16:33