Find $p>1$ that ${intlimits^p_1}frac{1}{x},mathrm{d}x={intlimits^p_1}lnleft(xright),mathrm{d}x$












3












$begingroup$


Find $p>1$ that $${displaystyleintlimits^p_1}dfrac{1}{x},mathrm{d}x={displaystyleintlimits^p_1}lnleft(xright),mathrm{d}x$$




begin{align*}
&{displaystyleint}dfrac{1}{x},mathrm{d}x=lnleft(mid x mid right) && vert text{general integral}
end{align*}
$F_1(x)=lnleft({mid x mid} right)+C$



begin{align*}
&{displaystyleint}1cdotlnleft(xright ),mathrm{d}x && vert 2. text{ with } f'=1, g=ln(x)\
&=xlnleft(xright)-{displaystyleint}1,mathrm{d}x\
&=xlnleft(xright)-x
end{align*}
$F_2(x)=xlnleft(xright)-x+C$




begin{align*}
left[lnleft(mid{x}midright)+Cright]^p_1&=left[ln({mid p mid})+Cright]-left[ln({mid 1 mid})+Cright]\
&=lnleft(pright)
end{align*}
begin{align*}
&left[xlnleft(x right)-x+Cright]^p_1=left[plnleft(p right)-p+Cright]-left[1lnleft(1right)-1+Cright]
end{align*}



It was already mentioned, that I had a typo. I corrected everything and contributed my own solution.






calculus analysis definite-integrals







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  • 2




    $begingroup$
    Why did you put $0$ instead of $1$?
    $endgroup$
    – metamorphy
    Dec 20 '18 at 10:41










  • $begingroup$
    Oh, I'm so dumb. Thanks, that was my mistake, it's 1 not 0.
    $endgroup$
    – Doesbaddel
    Dec 20 '18 at 10:41
















3












$begingroup$


Find $p>1$ that $${displaystyleintlimits^p_1}dfrac{1}{x},mathrm{d}x={displaystyleintlimits^p_1}lnleft(xright),mathrm{d}x$$




begin{align*}
&{displaystyleint}dfrac{1}{x},mathrm{d}x=lnleft(mid x mid right) && vert text{general integral}
end{align*}
$F_1(x)=lnleft({mid x mid} right)+C$



begin{align*}
&{displaystyleint}1cdotlnleft(xright ),mathrm{d}x && vert 2. text{ with } f'=1, g=ln(x)\
&=xlnleft(xright)-{displaystyleint}1,mathrm{d}x\
&=xlnleft(xright)-x
end{align*}
$F_2(x)=xlnleft(xright)-x+C$




begin{align*}
left[lnleft(mid{x}midright)+Cright]^p_1&=left[ln({mid p mid})+Cright]-left[ln({mid 1 mid})+Cright]\
&=lnleft(pright)
end{align*}
begin{align*}
&left[xlnleft(x right)-x+Cright]^p_1=left[plnleft(p right)-p+Cright]-left[1lnleft(1right)-1+Cright]
end{align*}



It was already mentioned, that I had a typo. I corrected everything and contributed my own solution.






calculus analysis definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Why did you put $0$ instead of $1$?
    $endgroup$
    – metamorphy
    Dec 20 '18 at 10:41










  • $begingroup$
    Oh, I'm so dumb. Thanks, that was my mistake, it's 1 not 0.
    $endgroup$
    – Doesbaddel
    Dec 20 '18 at 10:41














3












3








3


1



$begingroup$


Find $p>1$ that $${displaystyleintlimits^p_1}dfrac{1}{x},mathrm{d}x={displaystyleintlimits^p_1}lnleft(xright),mathrm{d}x$$




begin{align*}
&{displaystyleint}dfrac{1}{x},mathrm{d}x=lnleft(mid x mid right) && vert text{general integral}
end{align*}
$F_1(x)=lnleft({mid x mid} right)+C$



begin{align*}
&{displaystyleint}1cdotlnleft(xright ),mathrm{d}x && vert 2. text{ with } f'=1, g=ln(x)\
&=xlnleft(xright)-{displaystyleint}1,mathrm{d}x\
&=xlnleft(xright)-x
end{align*}
$F_2(x)=xlnleft(xright)-x+C$




begin{align*}
left[lnleft(mid{x}midright)+Cright]^p_1&=left[ln({mid p mid})+Cright]-left[ln({mid 1 mid})+Cright]\
&=lnleft(pright)
end{align*}
begin{align*}
&left[xlnleft(x right)-x+Cright]^p_1=left[plnleft(p right)-p+Cright]-left[1lnleft(1right)-1+Cright]
end{align*}



It was already mentioned, that I had a typo. I corrected everything and contributed my own solution.






calculus analysis definite-integrals







share|cite|improve this question











$endgroup$




Find $p>1$ that $${displaystyleintlimits^p_1}dfrac{1}{x},mathrm{d}x={displaystyleintlimits^p_1}lnleft(xright),mathrm{d}x$$




begin{align*}
&{displaystyleint}dfrac{1}{x},mathrm{d}x=lnleft(mid x mid right) && vert text{general integral}
end{align*}
$F_1(x)=lnleft({mid x mid} right)+C$



begin{align*}
&{displaystyleint}1cdotlnleft(xright ),mathrm{d}x && vert 2. text{ with } f'=1, g=ln(x)\
&=xlnleft(xright)-{displaystyleint}1,mathrm{d}x\
&=xlnleft(xright)-x
end{align*}
$F_2(x)=xlnleft(xright)-x+C$




begin{align*}
left[lnleft(mid{x}midright)+Cright]^p_1&=left[ln({mid p mid})+Cright]-left[ln({mid 1 mid})+Cright]\
&=lnleft(pright)
end{align*}
begin{align*}
&left[xlnleft(x right)-x+Cright]^p_1=left[plnleft(p right)-p+Cright]-left[1lnleft(1right)-1+Cright]
end{align*}



It was already mentioned, that I had a typo. I corrected everything and contributed my own solution.






calculus analysis definite-integrals




calculus analysis definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 12:08









Eff

11.5k21638




11.5k21638










asked Dec 20 '18 at 10:39









DoesbaddelDoesbaddel

33811




33811








  • 2




    $begingroup$
    Why did you put $0$ instead of $1$?
    $endgroup$
    – metamorphy
    Dec 20 '18 at 10:41










  • $begingroup$
    Oh, I'm so dumb. Thanks, that was my mistake, it's 1 not 0.
    $endgroup$
    – Doesbaddel
    Dec 20 '18 at 10:41














  • 2




    $begingroup$
    Why did you put $0$ instead of $1$?
    $endgroup$
    – metamorphy
    Dec 20 '18 at 10:41










  • $begingroup$
    Oh, I'm so dumb. Thanks, that was my mistake, it's 1 not 0.
    $endgroup$
    – Doesbaddel
    Dec 20 '18 at 10:41








2




2




$begingroup$
Why did you put $0$ instead of $1$?
$endgroup$
– metamorphy
Dec 20 '18 at 10:41




$begingroup$
Why did you put $0$ instead of $1$?
$endgroup$
– metamorphy
Dec 20 '18 at 10:41












$begingroup$
Oh, I'm so dumb. Thanks, that was my mistake, it's 1 not 0.
$endgroup$
– Doesbaddel
Dec 20 '18 at 10:41




$begingroup$
Oh, I'm so dumb. Thanks, that was my mistake, it's 1 not 0.
$endgroup$
– Doesbaddel
Dec 20 '18 at 10:41










2 Answers
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$begingroup$

$ int_1^p frac{1}{x} dx=int_1^p ln x dx iff ln p = p ln p-p+1 iff (1-p) ln p=1-p iff ln p=1 iff p=e.$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Find $p>1$ that
    $$
    {displaystyleintlimits^p_1}dfrac{1}{x},mathrm{d}x={displaystyleintlimits^p_1}lnleft(xright),mathrm{d}x
    $$




    begin{align*}
    &{displaystyleint}dfrac{1}{x},mathrm{d}x=lnleft(mid x mid right) && vert text{general integr}
    end{align*}
    $F_1(x)=lnleft({mid x mid} right)+C$



    begin{align*}
    &{displaystyleint}1cdotlnleft(xright ),mathrm{d}x && vert 2. text{ with } f'=1, g=ln(x)\
    &=xlnleft(xright)-{displaystyleint}1,mathrm{d}x\
    &=xlnleft(xright)-x
    end{align*}
    $F_2(x)=xlnleft(xright)-x+C$




    begin{align*}
    left[lnleft(mid{x}midright)+Cright]^p_1&=left[ln({mid p mid})+Cright]-left[ln({mid 0 mid})+Cright]\
    &=lnleft(pright)
    end{align*}
    begin{align*}
    left[xlnleft(x right)-x+Cright]^p_1&=left[plnleft(p right)-p+Cright]-left[1lnleft(1 right)-1+Cright]\
    &=left[plnleft(p right)-p+Cright]-[-1+C]\
    &=plnleft(p right)-p+1
    end{align*}
    Solve $lnleft(pright)=plnleft(p right)-p+1$ for $p$:
    begin{align*}
    &lnleft(pright)=plnleft(p right)-p+1&&vert -(1-p+pln(p)\
    &iff-1+p+ln(p)-pln(p)=0\
    &iff-(p-1)cdot (ln(p)-1)=0&& vert cdot (-1)\
    &iff (p-1)cdot (ln(p)-1)=0 && vert p>1\
    &iff (ln(p)-1)=0\
    &iff (ln(p))=1&& vert exp{()}\
    &iff underline{underline{p=e}}
    end{align*}






    share|cite|improve this answer











    $endgroup$













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      2 Answers
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      2 Answers
      2






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      2












      $begingroup$

      $ int_1^p frac{1}{x} dx=int_1^p ln x dx iff ln p = p ln p-p+1 iff (1-p) ln p=1-p iff ln p=1 iff p=e.$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $ int_1^p frac{1}{x} dx=int_1^p ln x dx iff ln p = p ln p-p+1 iff (1-p) ln p=1-p iff ln p=1 iff p=e.$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $ int_1^p frac{1}{x} dx=int_1^p ln x dx iff ln p = p ln p-p+1 iff (1-p) ln p=1-p iff ln p=1 iff p=e.$






          share|cite|improve this answer









          $endgroup$



          $ int_1^p frac{1}{x} dx=int_1^p ln x dx iff ln p = p ln p-p+1 iff (1-p) ln p=1-p iff ln p=1 iff p=e.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 11:05









          FredFred

          46.8k1848




          46.8k1848























              1












              $begingroup$

              Find $p>1$ that
              $$
              {displaystyleintlimits^p_1}dfrac{1}{x},mathrm{d}x={displaystyleintlimits^p_1}lnleft(xright),mathrm{d}x
              $$




              begin{align*}
              &{displaystyleint}dfrac{1}{x},mathrm{d}x=lnleft(mid x mid right) && vert text{general integr}
              end{align*}
              $F_1(x)=lnleft({mid x mid} right)+C$



              begin{align*}
              &{displaystyleint}1cdotlnleft(xright ),mathrm{d}x && vert 2. text{ with } f'=1, g=ln(x)\
              &=xlnleft(xright)-{displaystyleint}1,mathrm{d}x\
              &=xlnleft(xright)-x
              end{align*}
              $F_2(x)=xlnleft(xright)-x+C$




              begin{align*}
              left[lnleft(mid{x}midright)+Cright]^p_1&=left[ln({mid p mid})+Cright]-left[ln({mid 0 mid})+Cright]\
              &=lnleft(pright)
              end{align*}
              begin{align*}
              left[xlnleft(x right)-x+Cright]^p_1&=left[plnleft(p right)-p+Cright]-left[1lnleft(1 right)-1+Cright]\
              &=left[plnleft(p right)-p+Cright]-[-1+C]\
              &=plnleft(p right)-p+1
              end{align*}
              Solve $lnleft(pright)=plnleft(p right)-p+1$ for $p$:
              begin{align*}
              &lnleft(pright)=plnleft(p right)-p+1&&vert -(1-p+pln(p)\
              &iff-1+p+ln(p)-pln(p)=0\
              &iff-(p-1)cdot (ln(p)-1)=0&& vert cdot (-1)\
              &iff (p-1)cdot (ln(p)-1)=0 && vert p>1\
              &iff (ln(p)-1)=0\
              &iff (ln(p))=1&& vert exp{()}\
              &iff underline{underline{p=e}}
              end{align*}






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Find $p>1$ that
                $$
                {displaystyleintlimits^p_1}dfrac{1}{x},mathrm{d}x={displaystyleintlimits^p_1}lnleft(xright),mathrm{d}x
                $$




                begin{align*}
                &{displaystyleint}dfrac{1}{x},mathrm{d}x=lnleft(mid x mid right) && vert text{general integr}
                end{align*}
                $F_1(x)=lnleft({mid x mid} right)+C$



                begin{align*}
                &{displaystyleint}1cdotlnleft(xright ),mathrm{d}x && vert 2. text{ with } f'=1, g=ln(x)\
                &=xlnleft(xright)-{displaystyleint}1,mathrm{d}x\
                &=xlnleft(xright)-x
                end{align*}
                $F_2(x)=xlnleft(xright)-x+C$




                begin{align*}
                left[lnleft(mid{x}midright)+Cright]^p_1&=left[ln({mid p mid})+Cright]-left[ln({mid 0 mid})+Cright]\
                &=lnleft(pright)
                end{align*}
                begin{align*}
                left[xlnleft(x right)-x+Cright]^p_1&=left[plnleft(p right)-p+Cright]-left[1lnleft(1 right)-1+Cright]\
                &=left[plnleft(p right)-p+Cright]-[-1+C]\
                &=plnleft(p right)-p+1
                end{align*}
                Solve $lnleft(pright)=plnleft(p right)-p+1$ for $p$:
                begin{align*}
                &lnleft(pright)=plnleft(p right)-p+1&&vert -(1-p+pln(p)\
                &iff-1+p+ln(p)-pln(p)=0\
                &iff-(p-1)cdot (ln(p)-1)=0&& vert cdot (-1)\
                &iff (p-1)cdot (ln(p)-1)=0 && vert p>1\
                &iff (ln(p)-1)=0\
                &iff (ln(p))=1&& vert exp{()}\
                &iff underline{underline{p=e}}
                end{align*}






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Find $p>1$ that
                  $$
                  {displaystyleintlimits^p_1}dfrac{1}{x},mathrm{d}x={displaystyleintlimits^p_1}lnleft(xright),mathrm{d}x
                  $$




                  begin{align*}
                  &{displaystyleint}dfrac{1}{x},mathrm{d}x=lnleft(mid x mid right) && vert text{general integr}
                  end{align*}
                  $F_1(x)=lnleft({mid x mid} right)+C$



                  begin{align*}
                  &{displaystyleint}1cdotlnleft(xright ),mathrm{d}x && vert 2. text{ with } f'=1, g=ln(x)\
                  &=xlnleft(xright)-{displaystyleint}1,mathrm{d}x\
                  &=xlnleft(xright)-x
                  end{align*}
                  $F_2(x)=xlnleft(xright)-x+C$




                  begin{align*}
                  left[lnleft(mid{x}midright)+Cright]^p_1&=left[ln({mid p mid})+Cright]-left[ln({mid 0 mid})+Cright]\
                  &=lnleft(pright)
                  end{align*}
                  begin{align*}
                  left[xlnleft(x right)-x+Cright]^p_1&=left[plnleft(p right)-p+Cright]-left[1lnleft(1 right)-1+Cright]\
                  &=left[plnleft(p right)-p+Cright]-[-1+C]\
                  &=plnleft(p right)-p+1
                  end{align*}
                  Solve $lnleft(pright)=plnleft(p right)-p+1$ for $p$:
                  begin{align*}
                  &lnleft(pright)=plnleft(p right)-p+1&&vert -(1-p+pln(p)\
                  &iff-1+p+ln(p)-pln(p)=0\
                  &iff-(p-1)cdot (ln(p)-1)=0&& vert cdot (-1)\
                  &iff (p-1)cdot (ln(p)-1)=0 && vert p>1\
                  &iff (ln(p)-1)=0\
                  &iff (ln(p))=1&& vert exp{()}\
                  &iff underline{underline{p=e}}
                  end{align*}






                  share|cite|improve this answer











                  $endgroup$



                  Find $p>1$ that
                  $$
                  {displaystyleintlimits^p_1}dfrac{1}{x},mathrm{d}x={displaystyleintlimits^p_1}lnleft(xright),mathrm{d}x
                  $$




                  begin{align*}
                  &{displaystyleint}dfrac{1}{x},mathrm{d}x=lnleft(mid x mid right) && vert text{general integr}
                  end{align*}
                  $F_1(x)=lnleft({mid x mid} right)+C$



                  begin{align*}
                  &{displaystyleint}1cdotlnleft(xright ),mathrm{d}x && vert 2. text{ with } f'=1, g=ln(x)\
                  &=xlnleft(xright)-{displaystyleint}1,mathrm{d}x\
                  &=xlnleft(xright)-x
                  end{align*}
                  $F_2(x)=xlnleft(xright)-x+C$




                  begin{align*}
                  left[lnleft(mid{x}midright)+Cright]^p_1&=left[ln({mid p mid})+Cright]-left[ln({mid 0 mid})+Cright]\
                  &=lnleft(pright)
                  end{align*}
                  begin{align*}
                  left[xlnleft(x right)-x+Cright]^p_1&=left[plnleft(p right)-p+Cright]-left[1lnleft(1 right)-1+Cright]\
                  &=left[plnleft(p right)-p+Cright]-[-1+C]\
                  &=plnleft(p right)-p+1
                  end{align*}
                  Solve $lnleft(pright)=plnleft(p right)-p+1$ for $p$:
                  begin{align*}
                  &lnleft(pright)=plnleft(p right)-p+1&&vert -(1-p+pln(p)\
                  &iff-1+p+ln(p)-pln(p)=0\
                  &iff-(p-1)cdot (ln(p)-1)=0&& vert cdot (-1)\
                  &iff (p-1)cdot (ln(p)-1)=0 && vert p>1\
                  &iff (ln(p)-1)=0\
                  &iff (ln(p))=1&& vert exp{()}\
                  &iff underline{underline{p=e}}
                  end{align*}







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 21 '18 at 9:41

























                  answered Dec 20 '18 at 11:05









                  DoesbaddelDoesbaddel

                  33811




                  33811






























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