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












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

5411417




5411417












  • $begingroup$
    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














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1 Answer
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1 Answer
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active

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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

7,88511062




7,88511062












  • $begingroup$
    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




















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