Upper bound for the integral.

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

Is there an upper bound for the following integral in terms of $c$ and $A$ ?

$int_-infty^+inftyfracx+cx dx$,

where $c, Ain mathbb R$ and $Bin mathbb R^+$.

Here $A$ is fixed and $B$ is a parameter and can be chosen as large as possible. Simplest case is when $c=0$. In this case the integral is absolutely convergent (by choosing $B$ appropriately) and thus bounded by some absolute constant.

definite-integrals upper-lower-bounds

share|cite|improve this question

edited Jul 23 at 6:45

asked Jul 23 at 6:38

Praan

32

add a comment | 

up vote
0
down vote

favorite

Is there an upper bound for the following integral in terms of $c$ and $A$ ?

$int_-infty^+inftyfracx+cx dx$,

where $c, Ain mathbb R$ and $Bin mathbb R^+$.

Here $A$ is fixed and $B$ is a parameter and can be chosen as large as possible. Simplest case is when $c=0$. In this case the integral is absolutely convergent (by choosing $B$ appropriately) and thus bounded by some absolute constant.

definite-integrals upper-lower-bounds

share|cite|improve this question

edited Jul 23 at 6:45

asked Jul 23 at 6:38

Praan

32

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Is there an upper bound for the following integral in terms of $c$ and $A$ ?

$int_-infty^+inftyfracx+cx dx$,

where $c, Ain mathbb R$ and $Bin mathbb R^+$.

Here $A$ is fixed and $B$ is a parameter and can be chosen as large as possible. Simplest case is when $c=0$. In this case the integral is absolutely convergent (by choosing $B$ appropriately) and thus bounded by some absolute constant.

definite-integrals upper-lower-bounds

share|cite|improve this question

edited Jul 23 at 6:45

asked Jul 23 at 6:38

Praan

32

Is there an upper bound for the following integral in terms of $c$ and $A$ ?

$int_-infty^+inftyfracx+cx dx$,

where $c, Ain mathbb R$ and $Bin mathbb R^+$.

Here $A$ is fixed and $B$ is a parameter and can be chosen as large as possible. Simplest case is when $c=0$. In this case the integral is absolutely convergent (by choosing $B$ appropriately) and thus bounded by some absolute constant.

definite-integrals upper-lower-bounds

share|cite|improve this question

edited Jul 23 at 6:45

asked Jul 23 at 6:38

Praan

32

share|cite|improve this question

share|cite|improve this question

edited Jul 23 at 6:45

edited Jul 23 at 6:45

edited Jul 23 at 6:45

asked Jul 23 at 6:38

Praan

32

asked Jul 23 at 6:38

Praan

32

asked Jul 23 at 6:38

Praan

32

32

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

Hint:

It is possible to get an exact expression. Due to the fact that there is an arbitrary additive parameter $c$, the problem is not simpler than the evaluation of the indefinite integral. Then it is advisable to get rid of the absolute values by splitting the domain in the three intervals delimited by $0$ and $-c$.

So, with a shift of the variable it suffices to consider

$$intfrac(x+c)^ax^bdx.$$

When $a$ is an integer, you develop the binomial and end-up with a finite sum of powers. Otherwise, I don't think you can escape the incomplete Beta integral.

share|cite|improve this answer

answered Jul 23 at 7:00

Yves Daoust

111k665203

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860068%2fupper-bound-for-the-integral%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Hint:

It is possible to get an exact expression. Due to the fact that there is an arbitrary additive parameter $c$, the problem is not simpler than the evaluation of the indefinite integral. Then it is advisable to get rid of the absolute values by splitting the domain in the three intervals delimited by $0$ and $-c$.

So, with a shift of the variable it suffices to consider

$$intfrac(x+c)^ax^bdx.$$

When $a$ is an integer, you develop the binomial and end-up with a finite sum of powers. Otherwise, I don't think you can escape the incomplete Beta integral.

share|cite|improve this answer

answered Jul 23 at 7:00

Yves Daoust

111k665203

add a comment | 

up vote
0
down vote



accepted

Hint:

It is possible to get an exact expression. Due to the fact that there is an arbitrary additive parameter $c$, the problem is not simpler than the evaluation of the indefinite integral. Then it is advisable to get rid of the absolute values by splitting the domain in the three intervals delimited by $0$ and $-c$.

So, with a shift of the variable it suffices to consider

$$intfrac(x+c)^ax^bdx.$$

When $a$ is an integer, you develop the binomial and end-up with a finite sum of powers. Otherwise, I don't think you can escape the incomplete Beta integral.

share|cite|improve this answer

answered Jul 23 at 7:00

Yves Daoust

111k665203

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Hint:

It is possible to get an exact expression. Due to the fact that there is an arbitrary additive parameter $c$, the problem is not simpler than the evaluation of the indefinite integral. Then it is advisable to get rid of the absolute values by splitting the domain in the three intervals delimited by $0$ and $-c$.

So, with a shift of the variable it suffices to consider

$$intfrac(x+c)^ax^bdx.$$

When $a$ is an integer, you develop the binomial and end-up with a finite sum of powers. Otherwise, I don't think you can escape the incomplete Beta integral.

share|cite|improve this answer

answered Jul 23 at 7:00

Yves Daoust

111k665203

Hint:

It is possible to get an exact expression. Due to the fact that there is an arbitrary additive parameter $c$, the problem is not simpler than the evaluation of the indefinite integral. Then it is advisable to get rid of the absolute values by splitting the domain in the three intervals delimited by $0$ and $-c$.

So, with a shift of the variable it suffices to consider

$$intfrac(x+c)^ax^bdx.$$

When $a$ is an integer, you develop the binomial and end-up with a finite sum of powers. Otherwise, I don't think you can escape the incomplete Beta integral.

share|cite|improve this answer

answered Jul 23 at 7:00

Yves Daoust

111k665203

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 7:00

Yves Daoust

111k665203

answered Jul 23 at 7:00

Yves Daoust

111k665203

answered Jul 23 at 7:00

Yves Daoust

111k665203

111k665203

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860068%2fupper-bound-for-the-integral%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email