Fdtxjr

Does the reader need to like the PoV character?

What does Apple's new App Store requirement mean

Which Article Helped Get Rid of Technobabble in RPGs?

Why does this expression simplify as such?

Non-trope happy ending?

Why is so much work done on numerical verification of the Riemann Hypothesis?

Can you use Vicious Mockery to win an argument or gain favours?

How can ping know if my host is down

What is the highest possible scrabble score for placing a single tile

Why should universal income be universal?

How would you translate "more" for use as an interface button?

Is there a nicer/politer/more positive alternative for "negates"?

Can I cause damage to electrical appliances by unplugging them when they are turned on?

Does the Linux kernel need a file system to run?

C++ copy constructor called at return

How to convince somebody that he is fit for something else, but not this job?

Why do ¬, ∀ and ∃ have the same precedence?

Is it necessary to use pronouns with the verb "essere"?

How to get directions in deep space?

Is there a way to have vectors outlined in a Vector Plot?

Does an advisor owe his/her student anything? Will an advisor keep a PhD student only out of pity?

Microchip documentation does not label CAN buss pins on micro controller pinout diagram

I found an audio circuit and I built it just fine, but I find it a bit too quiet. How do I amplify the output so that it is a bit louder?

Is this toilet slogan correct usage of the English language?

Why is $int_0^1 log x $ computable?

How do I calculate $lim_xrightarrow 0 xln x$Problem understanding Integration questionContour integration of log over polynomial with fractional powerIntegral $int_0^inftyBig[logleft(1+x^2right)-psileft(1+x^2right)Big]dx$Deriving Separate Forms of the Error FunctionEvaluate the integral $ int_0^infty r^2 e^-r^2/2sigma^2 dr$Evaluate $int_0^inftyfraclog^2 xe^x^2mathrmdx$.Should I take the value of $int_0^2pitan^5 (x) dx$ as $0$ or as undefined?Definite Integral involving natural logExplain why $int_0^pi/2log(sin(x))textdx=-int_0^pi/2xcot(x)$dxHow to evaluate $int_-infty^inftyfracxarctanfrac1x log(1+x^2)1+x^2dx$

2

$begingroup$

While evaluating $$lim_n to inftyleft(fracn!nright)^1/n $$

The integral turns out to be

$$int_0^1 log x = big[xlog xbig]_0^1 - big[x big]_0^1 $$

The second term will -1 how ever the first term will be
$$I_1=1timeslog 1 -0times log 0$$

My text book has taken the value of $0log 0$ as $0$. However isn't $log0$ undefined?

definite-integrals

share|cite|improve this question

edited Mar 14 at 17:08

Dr. Sonnhard Graubner

77.9k42866

asked Mar 14 at 16:54

CaptainQuestionCaptainQuestion

1749

$endgroup$

add a comment | 

2

$begingroup$

While evaluating $$lim_n to inftyleft(fracn!nright)^1/n $$

The integral turns out to be

$$int_0^1 log x = big[xlog xbig]_0^1 - big[x big]_0^1 $$

The second term will -1 how ever the first term will be
$$I_1=1timeslog 1 -0times log 0$$

My text book has taken the value of $0log 0$ as $0$. However isn't $log0$ undefined?

definite-integrals

share|cite|improve this question

edited Mar 14 at 17:08

Dr. Sonnhard Graubner

77.9k42866

asked Mar 14 at 16:54

CaptainQuestionCaptainQuestion

1749

$endgroup$

add a comment | 

$begingroup$

While evaluating $$lim_n to inftyleft(fracn!nright)^1/n $$

The integral turns out to be

$$int_0^1 log x = big[xlog xbig]_0^1 - big[x big]_0^1 $$

The second term will -1 how ever the first term will be
$$I_1=1timeslog 1 -0times log 0$$

My text book has taken the value of $0log 0$ as $0$. However isn't $log0$ undefined?

definite-integrals

share|cite|improve this question

edited Mar 14 at 17:08

Dr. Sonnhard Graubner

77.9k42866

asked Mar 14 at 16:54

CaptainQuestionCaptainQuestion

1749

$endgroup$

share|cite|improve this question

edited Mar 14 at 17:08

Dr. Sonnhard Graubner

77.9k42866

asked Mar 14 at 16:54

CaptainQuestionCaptainQuestion

1749

share|cite|improve this question

edited Mar 14 at 17:08

Dr. Sonnhard Graubner

77.9k42866

asked Mar 14 at 16:54

CaptainQuestionCaptainQuestion

1749

edited Mar 14 at 17:08

Dr. Sonnhard Graubner

77.9k42866

edited Mar 14 at 17:08

Dr. Sonnhard Graubner

77.9k42866

asked Mar 14 at 16:54

CaptainQuestionCaptainQuestion

1749

asked Mar 14 at 16:54

CaptainQuestionCaptainQuestion

1749

2 Answers
2

active

oldest

votes

5

$begingroup$

Since $log(0)$ is undefined, you have to take the one-sided (!) limit:
$$
L
:= lim_x searrow 0 x cdot log(x)
= 0.
$$
This is zero, which you can obtain by using L'Hospitals rule:
beginalign
L
= lim_x searrow 0 fraclog(x)frac1x
oversettextL'H= lim_x searrow 0 fracfrac1x- frac1x^2
= lim_x searrow 0 - fracx^2x
= lim_x searrow 0 -x
= 0.
endalign

---

Basically, since log(x) isn't defined for one of the bounds of integration, I would always begin by writing
beginalign
int_0^1 log(x) dx
& = lim_a searrow 0 int_a^1 log(x) dx
= lim_a searrow 0 big[ x left( log(x) - 1right) big]_a^1 \
& = lim_a searrow 0 1 ( 0 - 1) - a left( log(a) - 1right)
= -1 - lim_x searrow 0 x left( log(x) - 1right)
endalign

share|cite|improve this answer

edited Mar 14 at 17:04

answered Mar 14 at 16:56

Viktor GlombikViktor Glombik

1,2032528

$endgroup$

add a comment | 

6

$begingroup$

Rather than (ab)using L'Hôpital's rule, you should know (perhaps once and for all by L'Hôpital, but there are other ways) that as $xtoinfty$ exponential functions beat out any power of $x$. It follows that as $xtoinfty$ any power of the logarithm loses to $x$, i.e.,
$$lim_xtoinfty frac(log x)^nx = 0.$$
Now, note that
$$lim_xto 0^+ xlog x = lim_utoinfty frac1ulogbig(frac1ubig) = lim_utoinftyfrac-log(u)u = 0.$$

share|cite|improve this answer

answered Mar 14 at 17:02

Ted ShifrinTed Shifrin

64.5k44692

$endgroup$

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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
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

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3148244%2fwhy-is-int-01-log-x-computable%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

Since $log(0)$ is undefined, you have to take the one-sided (!) limit:
$$
L
:= lim_x searrow 0 x cdot log(x)
= 0.
$$
This is zero, which you can obtain by using L'Hospitals rule:
beginalign
L
= lim_x searrow 0 fraclog(x)frac1x
oversettextL'H= lim_x searrow 0 fracfrac1x- frac1x^2
= lim_x searrow 0 - fracx^2x
= lim_x searrow 0 -x
= 0.
endalign

---

Basically, since log(x) isn't defined for one of the bounds of integration, I would always begin by writing
beginalign
int_0^1 log(x) dx
& = lim_a searrow 0 int_a^1 log(x) dx
= lim_a searrow 0 big[ x left( log(x) - 1right) big]_a^1 \
& = lim_a searrow 0 1 ( 0 - 1) - a left( log(a) - 1right)
= -1 - lim_x searrow 0 x left( log(x) - 1right)
endalign

share|cite|improve this answer

edited Mar 14 at 17:04

answered Mar 14 at 16:56

Viktor GlombikViktor Glombik

1,2032528

$endgroup$

add a comment | 

5

$begingroup$

Since $log(0)$ is undefined, you have to take the one-sided (!) limit:
$$
L
:= lim_x searrow 0 x cdot log(x)
= 0.
$$
This is zero, which you can obtain by using L'Hospitals rule:
beginalign
L
= lim_x searrow 0 fraclog(x)frac1x
oversettextL'H= lim_x searrow 0 fracfrac1x- frac1x^2
= lim_x searrow 0 - fracx^2x
= lim_x searrow 0 -x
= 0.
endalign

---

Basically, since log(x) isn't defined for one of the bounds of integration, I would always begin by writing
beginalign
int_0^1 log(x) dx
& = lim_a searrow 0 int_a^1 log(x) dx
= lim_a searrow 0 big[ x left( log(x) - 1right) big]_a^1 \
& = lim_a searrow 0 1 ( 0 - 1) - a left( log(a) - 1right)
= -1 - lim_x searrow 0 x left( log(x) - 1right)
endalign

share|cite|improve this answer

edited Mar 14 at 17:04

answered Mar 14 at 16:56

Viktor GlombikViktor Glombik

1,2032528

$endgroup$

add a comment | 

$begingroup$

Since $log(0)$ is undefined, you have to take the one-sided (!) limit:
$$
L
:= lim_x searrow 0 x cdot log(x)
= 0.
$$
This is zero, which you can obtain by using L'Hospitals rule:
beginalign
L
= lim_x searrow 0 fraclog(x)frac1x
oversettextL'H= lim_x searrow 0 fracfrac1x- frac1x^2
= lim_x searrow 0 - fracx^2x
= lim_x searrow 0 -x
= 0.
endalign

---

Basically, since log(x) isn't defined for one of the bounds of integration, I would always begin by writing
beginalign
int_0^1 log(x) dx
& = lim_a searrow 0 int_a^1 log(x) dx
= lim_a searrow 0 big[ x left( log(x) - 1right) big]_a^1 \
& = lim_a searrow 0 1 ( 0 - 1) - a left( log(a) - 1right)
= -1 - lim_x searrow 0 x left( log(x) - 1right)
endalign

share|cite|improve this answer

edited Mar 14 at 17:04

answered Mar 14 at 16:56

Viktor GlombikViktor Glombik

1,2032528

$endgroup$

share|cite|improve this answer

edited Mar 14 at 17:04

answered Mar 14 at 16:56

Viktor GlombikViktor Glombik

1,2032528

answered Mar 14 at 16:56

Viktor GlombikViktor Glombik

1,2032528

answered Mar 14 at 16:56

Viktor GlombikViktor Glombik

1,2032528

6

$begingroup$

Rather than (ab)using L'Hôpital's rule, you should know (perhaps once and for all by L'Hôpital, but there are other ways) that as $xtoinfty$ exponential functions beat out any power of $x$. It follows that as $xtoinfty$ any power of the logarithm loses to $x$, i.e.,
$$lim_xtoinfty frac(log x)^nx = 0.$$
Now, note that
$$lim_xto 0^+ xlog x = lim_utoinfty frac1ulogbig(frac1ubig) = lim_utoinftyfrac-log(u)u = 0.$$

share|cite|improve this answer

answered Mar 14 at 17:02

Ted ShifrinTed Shifrin

64.5k44692

$endgroup$

add a comment | 

6

$begingroup$

Rather than (ab)using L'Hôpital's rule, you should know (perhaps once and for all by L'Hôpital, but there are other ways) that as $xtoinfty$ exponential functions beat out any power of $x$. It follows that as $xtoinfty$ any power of the logarithm loses to $x$, i.e.,
$$lim_xtoinfty frac(log x)^nx = 0.$$
Now, note that
$$lim_xto 0^+ xlog x = lim_utoinfty frac1ulogbig(frac1ubig) = lim_utoinftyfrac-log(u)u = 0.$$

share|cite|improve this answer

answered Mar 14 at 17:02

Ted ShifrinTed Shifrin

64.5k44692

$endgroup$

add a comment | 

$begingroup$

Rather than (ab)using L'Hôpital's rule, you should know (perhaps once and for all by L'Hôpital, but there are other ways) that as $xtoinfty$ exponential functions beat out any power of $x$. It follows that as $xtoinfty$ any power of the logarithm loses to $x$, i.e.,
$$lim_xtoinfty frac(log x)^nx = 0.$$
Now, note that
$$lim_xto 0^+ xlog x = lim_utoinfty frac1ulogbig(frac1ubig) = lim_utoinftyfrac-log(u)u = 0.$$

share|cite|improve this answer

answered Mar 14 at 17:02

Ted ShifrinTed Shifrin

64.5k44692

$endgroup$

share|cite|improve this answer

answered Mar 14 at 17:02

Ted ShifrinTed Shifrin

64.5k44692

answered Mar 14 at 17:02

Ted ShifrinTed Shifrin

64.5k44692

answered Mar 14 at 17:02

Ted ShifrinTed Shifrin

64.5k44692