$int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x = 0$?












2












$begingroup$



Does
$$int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x = 0$$
for all real numbers $a > 0$ and $b < 2$?




I came across this conjecture by showing its validity for the positive integer values of $a$ only.



To derive the result for positive integer $a$, make the substitution $u=frac1x$ on
$$ int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x $$



and we get
$$ int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x \
implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0
$$



From here, we can use Leibniz's rule of integration (differentiating with respect to $b$) $n$ times to retrieve
$$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x = 0$$



Yet I'm guessing that it is also valid for any real positive value of $a$ from numerical evidence. Complex methods are welcome but I won't really be familiar with them, so I would prefer sticking to real methods.






calculus integration definite-integrals improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I am not quite sure: How do you get from $$int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x$$ to $$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x$$? With respect to which parameter do you differentiate?
    $endgroup$
    – mrtaurho
    Dec 30 '18 at 14:40










  • $begingroup$
    Differentiate with respect to $b$.
    $endgroup$
    – Mint
    Dec 30 '18 at 14:41






  • 1




    $begingroup$
    I would include this detail hence, at least to me, it was not clear at all.
    $endgroup$
    – mrtaurho
    Dec 30 '18 at 14:45






  • 1




    $begingroup$
    Same proof as before.
    $endgroup$
    – Jack D'Aurizio
    Dec 30 '18 at 15:00
















2












$begingroup$



Does
$$int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x = 0$$
for all real numbers $a > 0$ and $b < 2$?




I came across this conjecture by showing its validity for the positive integer values of $a$ only.



To derive the result for positive integer $a$, make the substitution $u=frac1x$ on
$$ int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x $$



and we get
$$ int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x \
implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0
$$



From here, we can use Leibniz's rule of integration (differentiating with respect to $b$) $n$ times to retrieve
$$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x = 0$$



Yet I'm guessing that it is also valid for any real positive value of $a$ from numerical evidence. Complex methods are welcome but I won't really be familiar with them, so I would prefer sticking to real methods.






calculus integration definite-integrals improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I am not quite sure: How do you get from $$int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x$$ to $$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x$$? With respect to which parameter do you differentiate?
    $endgroup$
    – mrtaurho
    Dec 30 '18 at 14:40










  • $begingroup$
    Differentiate with respect to $b$.
    $endgroup$
    – Mint
    Dec 30 '18 at 14:41






  • 1




    $begingroup$
    I would include this detail hence, at least to me, it was not clear at all.
    $endgroup$
    – mrtaurho
    Dec 30 '18 at 14:45






  • 1




    $begingroup$
    Same proof as before.
    $endgroup$
    – Jack D'Aurizio
    Dec 30 '18 at 15:00














2












2








2





$begingroup$



Does
$$int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x = 0$$
for all real numbers $a > 0$ and $b < 2$?




I came across this conjecture by showing its validity for the positive integer values of $a$ only.



To derive the result for positive integer $a$, make the substitution $u=frac1x$ on
$$ int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x $$



and we get
$$ int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x \
implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0
$$



From here, we can use Leibniz's rule of integration (differentiating with respect to $b$) $n$ times to retrieve
$$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x = 0$$



Yet I'm guessing that it is also valid for any real positive value of $a$ from numerical evidence. Complex methods are welcome but I won't really be familiar with them, so I would prefer sticking to real methods.






calculus integration definite-integrals improper-integrals







share|cite|improve this question











$endgroup$





Does
$$int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x = 0$$
for all real numbers $a > 0$ and $b < 2$?




I came across this conjecture by showing its validity for the positive integer values of $a$ only.



To derive the result for positive integer $a$, make the substitution $u=frac1x$ on
$$ int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x $$



and we get
$$ int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x \
implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0
$$



From here, we can use Leibniz's rule of integration (differentiating with respect to $b$) $n$ times to retrieve
$$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x = 0$$



Yet I'm guessing that it is also valid for any real positive value of $a$ from numerical evidence. Complex methods are welcome but I won't really be familiar with them, so I would prefer sticking to real methods.






calculus integration definite-integrals improper-integrals




calculus integration definite-integrals improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 14:56







Mint

















asked Dec 30 '18 at 14:35









MintMint

5311417




5311417












  • $begingroup$
    I am not quite sure: How do you get from $$int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x$$ to $$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x$$? With respect to which parameter do you differentiate?
    $endgroup$
    – mrtaurho
    Dec 30 '18 at 14:40










  • $begingroup$
    Differentiate with respect to $b$.
    $endgroup$
    – Mint
    Dec 30 '18 at 14:41






  • 1




    $begingroup$
    I would include this detail hence, at least to me, it was not clear at all.
    $endgroup$
    – mrtaurho
    Dec 30 '18 at 14:45






  • 1




    $begingroup$
    Same proof as before.
    $endgroup$
    – Jack D'Aurizio
    Dec 30 '18 at 15:00


















  • $begingroup$
    I am not quite sure: How do you get from $$int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x$$ to $$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x$$? With respect to which parameter do you differentiate?
    $endgroup$
    – mrtaurho
    Dec 30 '18 at 14:40










  • $begingroup$
    Differentiate with respect to $b$.
    $endgroup$
    – Mint
    Dec 30 '18 at 14:41






  • 1




    $begingroup$
    I would include this detail hence, at least to me, it was not clear at all.
    $endgroup$
    – mrtaurho
    Dec 30 '18 at 14:45






  • 1




    $begingroup$
    Same proof as before.
    $endgroup$
    – Jack D'Aurizio
    Dec 30 '18 at 15:00
















$begingroup$
I am not quite sure: How do you get from $$int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x$$ to $$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x$$? With respect to which parameter do you differentiate?
$endgroup$
– mrtaurho
Dec 30 '18 at 14:40




$begingroup$
I am not quite sure: How do you get from $$int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x$$ to $$int_0^infty frac{x^nln(x)}{(x^2-bx+1)^{n+1}}{rm d}x$$? With respect to which parameter do you differentiate?
$endgroup$
– mrtaurho
Dec 30 '18 at 14:40












$begingroup$
Differentiate with respect to $b$.
$endgroup$
– Mint
Dec 30 '18 at 14:41




$begingroup$
Differentiate with respect to $b$.
$endgroup$
– Mint
Dec 30 '18 at 14:41




1




1




$begingroup$
I would include this detail hence, at least to me, it was not clear at all.
$endgroup$
– mrtaurho
Dec 30 '18 at 14:45




$begingroup$
I would include this detail hence, at least to me, it was not clear at all.
$endgroup$
– mrtaurho
Dec 30 '18 at 14:45




1




1




$begingroup$
Same proof as before.
$endgroup$
– Jack D'Aurizio
Dec 30 '18 at 15:00




$begingroup$
Same proof as before.
$endgroup$
– Jack D'Aurizio
Dec 30 '18 at 15:00










1 Answer
1






active

oldest

votes


















3












$begingroup$

Of course it does. Just substitute $displaystyle{x=frac{1}{t}}$ to get:
$$I=int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x=int_infty^0 frac{1}{t^a} frac{lnleft(frac{1}{t}right)}{left(frac{1}{t^2}-frac{b}{t}+1right)^{a+1}}frac{-dt}{t^2}$$
$$=int_0^infty frac{t^{2a+2}lnleft(frac{1}{t}right)}{left(1-bt+t^2right)^{a+1}}frac{dt}{t^{a+2}}=int_0^infty frac{t^alnleft(frac{1}{t}right)}{(t^2-bt+1)^{a+1}}dt=-I$$
$$I=-IRightarrow I=0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Omg I feel so dumb, thank you so much. Standard substitutions slip by me occasionally...
    $endgroup$
    – Mint
    Dec 30 '18 at 14:51






  • 2




    $begingroup$
    I like your derivation in your question! A small typo:$$int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - color{red}k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0$$
    $endgroup$
    – Zacky
    Dec 30 '18 at 14:53













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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056877%2fint-0-infty-fracxa-lnxx2-bx1a1-rm-dx-0%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Of course it does. Just substitute $displaystyle{x=frac{1}{t}}$ to get:
$$I=int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x=int_infty^0 frac{1}{t^a} frac{lnleft(frac{1}{t}right)}{left(frac{1}{t^2}-frac{b}{t}+1right)^{a+1}}frac{-dt}{t^2}$$
$$=int_0^infty frac{t^{2a+2}lnleft(frac{1}{t}right)}{left(1-bt+t^2right)^{a+1}}frac{dt}{t^{a+2}}=int_0^infty frac{t^alnleft(frac{1}{t}right)}{(t^2-bt+1)^{a+1}}dt=-I$$
$$I=-IRightarrow I=0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Omg I feel so dumb, thank you so much. Standard substitutions slip by me occasionally...
    $endgroup$
    – Mint
    Dec 30 '18 at 14:51






  • 2




    $begingroup$
    I like your derivation in your question! A small typo:$$int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - color{red}k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0$$
    $endgroup$
    – Zacky
    Dec 30 '18 at 14:53


















3












$begingroup$

Of course it does. Just substitute $displaystyle{x=frac{1}{t}}$ to get:
$$I=int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x=int_infty^0 frac{1}{t^a} frac{lnleft(frac{1}{t}right)}{left(frac{1}{t^2}-frac{b}{t}+1right)^{a+1}}frac{-dt}{t^2}$$
$$=int_0^infty frac{t^{2a+2}lnleft(frac{1}{t}right)}{left(1-bt+t^2right)^{a+1}}frac{dt}{t^{a+2}}=int_0^infty frac{t^alnleft(frac{1}{t}right)}{(t^2-bt+1)^{a+1}}dt=-I$$
$$I=-IRightarrow I=0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Omg I feel so dumb, thank you so much. Standard substitutions slip by me occasionally...
    $endgroup$
    – Mint
    Dec 30 '18 at 14:51






  • 2




    $begingroup$
    I like your derivation in your question! A small typo:$$int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - color{red}k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0$$
    $endgroup$
    – Zacky
    Dec 30 '18 at 14:53
















3












3








3





$begingroup$

Of course it does. Just substitute $displaystyle{x=frac{1}{t}}$ to get:
$$I=int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x=int_infty^0 frac{1}{t^a} frac{lnleft(frac{1}{t}right)}{left(frac{1}{t^2}-frac{b}{t}+1right)^{a+1}}frac{-dt}{t^2}$$
$$=int_0^infty frac{t^{2a+2}lnleft(frac{1}{t}right)}{left(1-bt+t^2right)^{a+1}}frac{dt}{t^{a+2}}=int_0^infty frac{t^alnleft(frac{1}{t}right)}{(t^2-bt+1)^{a+1}}dt=-I$$
$$I=-IRightarrow I=0$$






share|cite|improve this answer











$endgroup$



Of course it does. Just substitute $displaystyle{x=frac{1}{t}}$ to get:
$$I=int_0^infty frac{x^aln(x)}{(x^2-bx+1)^{a+1}}{rm d}x=int_infty^0 frac{1}{t^a} frac{lnleft(frac{1}{t}right)}{left(frac{1}{t^2}-frac{b}{t}+1right)^{a+1}}frac{-dt}{t^2}$$
$$=int_0^infty frac{t^{2a+2}lnleft(frac{1}{t}right)}{left(1-bt+t^2right)^{a+1}}frac{dt}{t^{a+2}}=int_0^infty frac{t^alnleft(frac{1}{t}right)}{(t^2-bt+1)^{a+1}}dt=-I$$
$$I=-IRightarrow I=0$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 15:02

























answered Dec 30 '18 at 14:46









ZackyZacky

7,1501961




7,1501961












  • $begingroup$
    Omg I feel so dumb, thank you so much. Standard substitutions slip by me occasionally...
    $endgroup$
    – Mint
    Dec 30 '18 at 14:51






  • 2




    $begingroup$
    I like your derivation in your question! A small typo:$$int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - color{red}k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0$$
    $endgroup$
    – Zacky
    Dec 30 '18 at 14:53




















  • $begingroup$
    Omg I feel so dumb, thank you so much. Standard substitutions slip by me occasionally...
    $endgroup$
    – Mint
    Dec 30 '18 at 14:51






  • 2




    $begingroup$
    I like your derivation in your question! A small typo:$$int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - color{red}k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0$$
    $endgroup$
    – Zacky
    Dec 30 '18 at 14:53


















$begingroup$
Omg I feel so dumb, thank you so much. Standard substitutions slip by me occasionally...
$endgroup$
– Mint
Dec 30 '18 at 14:51




$begingroup$
Omg I feel so dumb, thank you so much. Standard substitutions slip by me occasionally...
$endgroup$
– Mint
Dec 30 '18 at 14:51




2




2




$begingroup$
I like your derivation in your question! A small typo:$$int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - color{red}k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0$$
$endgroup$
– Zacky
Dec 30 '18 at 14:53






$begingroup$
I like your derivation in your question! A small typo:$$int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x = int_0^infty frac{ln(1+x^k)}{x^2-bx+1}{rm d}x - color{red}k int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x implies int_0^infty frac{ln(x)}{x^2-bx+1}{rm d}x = 0$$
$endgroup$
– Zacky
Dec 30 '18 at 14:53




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056877%2fint-0-infty-fracxa-lnxx2-bx1a1-rm-dx-0%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more