Improper Integration $int_0^inftyfracln(xpi^x),dxx$ converges or not.

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I was trying to find whether the $int_0^inftyfracln(xpi^x),dxx$ converges or not. And after integrating I got



$lim_xtoinfty(frac12ln^2(x)+xlnpi)-lim_xto0(frac12ln^2(x)+xlnpi)$. (Which I will define as A)



and I was lost there. Because I got $infty$-$infty$. Does that mean the integral diverges to something that is indeterminable.(not $infty$ and -$infty$)



Also I asked a question concerning the expression above here
L'hospital for inf-inf



And found that according to a fellow name gimusi, $$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$ (which I will define as B)



is, however calculable. And I calculated it to be approaching infinity. Does that mean the integral diverges (to infinity)? I am lost now.



Questions Summary:



  1. Does A=B (doesn't seem so)

  2. Does the integral converges or diverges? (And how to figure that out)









share|cite|improve this question



























    up vote
    4
    down vote

    favorite
    1












    I was trying to find whether the $int_0^inftyfracln(xpi^x),dxx$ converges or not. And after integrating I got



    $lim_xtoinfty(frac12ln^2(x)+xlnpi)-lim_xto0(frac12ln^2(x)+xlnpi)$. (Which I will define as A)



    and I was lost there. Because I got $infty$-$infty$. Does that mean the integral diverges to something that is indeterminable.(not $infty$ and -$infty$)



    Also I asked a question concerning the expression above here
    L'hospital for inf-inf



    And found that according to a fellow name gimusi, $$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$ (which I will define as B)



    is, however calculable. And I calculated it to be approaching infinity. Does that mean the integral diverges (to infinity)? I am lost now.



    Questions Summary:



    1. Does A=B (doesn't seem so)

    2. Does the integral converges or diverges? (And how to figure that out)









    share|cite|improve this question

























      up vote
      4
      down vote

      favorite
      1









      up vote
      4
      down vote

      favorite
      1






      1





      I was trying to find whether the $int_0^inftyfracln(xpi^x),dxx$ converges or not. And after integrating I got



      $lim_xtoinfty(frac12ln^2(x)+xlnpi)-lim_xto0(frac12ln^2(x)+xlnpi)$. (Which I will define as A)



      and I was lost there. Because I got $infty$-$infty$. Does that mean the integral diverges to something that is indeterminable.(not $infty$ and -$infty$)



      Also I asked a question concerning the expression above here
      L'hospital for inf-inf



      And found that according to a fellow name gimusi, $$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$ (which I will define as B)



      is, however calculable. And I calculated it to be approaching infinity. Does that mean the integral diverges (to infinity)? I am lost now.



      Questions Summary:



      1. Does A=B (doesn't seem so)

      2. Does the integral converges or diverges? (And how to figure that out)









      share|cite|improve this question















      I was trying to find whether the $int_0^inftyfracln(xpi^x),dxx$ converges or not. And after integrating I got



      $lim_xtoinfty(frac12ln^2(x)+xlnpi)-lim_xto0(frac12ln^2(x)+xlnpi)$. (Which I will define as A)



      and I was lost there. Because I got $infty$-$infty$. Does that mean the integral diverges to something that is indeterminable.(not $infty$ and -$infty$)



      Also I asked a question concerning the expression above here
      L'hospital for inf-inf



      And found that according to a fellow name gimusi, $$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$ (which I will define as B)



      is, however calculable. And I calculated it to be approaching infinity. Does that mean the integral diverges (to infinity)? I am lost now.



      Questions Summary:



      1. Does A=B (doesn't seem so)

      2. Does the integral converges or diverges? (And how to figure that out)






      calculus improper-integrals






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      edited Sep 9 at 9:02









      Bernard

      112k636105




      112k636105










      asked Sep 9 at 8:48









      JoisBack

      1355




      1355




















          1 Answer
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          up vote
          6
          down vote



          accepted










          We have that



          $$int_0^inftyfracln(xpi^x)xdx
          =int_0^inftyfracln x +xln pi xdx
          =int_0^inftyfracln xxdx+int_0^inftyln pi ,dx$$



          and we can conclude here since both integrals diverge.



          Note that for the limit we have



          $$left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(1/x)-(1/x)lnpi$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(x)-(1/x)lnpi$$



          $$xlnpi -(1/x)lnpi to infty$$



          but it can't be used to evaluate the integral indeed for the improper integral we need to consider for $a>0$ the two integrals



          $$int_0^inftyfracln(xpi^x)xdx=int_0^a fracln(xpi^x)xdx+int_a^inftyfracln(xpi^x)xdx$$



          and the two limits at $0$ and $infty$ for the two integrals are independent and must be evaluated separetely.






          share|cite|improve this answer






















          • Big thanks to your effort for explaining. However, can you pls explain why integral of lnx/x from x =0 to x=inf diverges? I mean I can understand that from x=1 to inf, it diverges. But here there are both plus and minus infinity at . Visually, I plotted the graph of lnx/x on desmos and found that at 0<x<1 the area is approaching minus infinity whereas when x>1 the area is approaching infinity? I know that it's coming back to the same question of infinity minus infinity, but I still don't understand why. Also the transition from the expression in line 3 to line 4 seems incorrect to me.
            – JoisBack
            Sep 9 at 9:24











          • It suffices note that $int_1^inftyfraclog xx$ diverges to conclude that the whole integral diverges. Thiems topic was discussed many times here on MSE, I try to find some link to those discussions.
            – gimusi
            Sep 9 at 16:06







          • 1




            For the second issue I’ve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
            – gimusi
            Sep 9 at 16:08










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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          6
          down vote



          accepted










          We have that



          $$int_0^inftyfracln(xpi^x)xdx
          =int_0^inftyfracln x +xln pi xdx
          =int_0^inftyfracln xxdx+int_0^inftyln pi ,dx$$



          and we can conclude here since both integrals diverge.



          Note that for the limit we have



          $$left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(1/x)-(1/x)lnpi$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(x)-(1/x)lnpi$$



          $$xlnpi -(1/x)lnpi to infty$$



          but it can't be used to evaluate the integral indeed for the improper integral we need to consider for $a>0$ the two integrals



          $$int_0^inftyfracln(xpi^x)xdx=int_0^a fracln(xpi^x)xdx+int_a^inftyfracln(xpi^x)xdx$$



          and the two limits at $0$ and $infty$ for the two integrals are independent and must be evaluated separetely.






          share|cite|improve this answer






















          • Big thanks to your effort for explaining. However, can you pls explain why integral of lnx/x from x =0 to x=inf diverges? I mean I can understand that from x=1 to inf, it diverges. But here there are both plus and minus infinity at . Visually, I plotted the graph of lnx/x on desmos and found that at 0<x<1 the area is approaching minus infinity whereas when x>1 the area is approaching infinity? I know that it's coming back to the same question of infinity minus infinity, but I still don't understand why. Also the transition from the expression in line 3 to line 4 seems incorrect to me.
            – JoisBack
            Sep 9 at 9:24











          • It suffices note that $int_1^inftyfraclog xx$ diverges to conclude that the whole integral diverges. Thiems topic was discussed many times here on MSE, I try to find some link to those discussions.
            – gimusi
            Sep 9 at 16:06







          • 1




            For the second issue I’ve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
            – gimusi
            Sep 9 at 16:08














          up vote
          6
          down vote



          accepted










          We have that



          $$int_0^inftyfracln(xpi^x)xdx
          =int_0^inftyfracln x +xln pi xdx
          =int_0^inftyfracln xxdx+int_0^inftyln pi ,dx$$



          and we can conclude here since both integrals diverge.



          Note that for the limit we have



          $$left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(1/x)-(1/x)lnpi$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(x)-(1/x)lnpi$$



          $$xlnpi -(1/x)lnpi to infty$$



          but it can't be used to evaluate the integral indeed for the improper integral we need to consider for $a>0$ the two integrals



          $$int_0^inftyfracln(xpi^x)xdx=int_0^a fracln(xpi^x)xdx+int_a^inftyfracln(xpi^x)xdx$$



          and the two limits at $0$ and $infty$ for the two integrals are independent and must be evaluated separetely.






          share|cite|improve this answer






















          • Big thanks to your effort for explaining. However, can you pls explain why integral of lnx/x from x =0 to x=inf diverges? I mean I can understand that from x=1 to inf, it diverges. But here there are both plus and minus infinity at . Visually, I plotted the graph of lnx/x on desmos and found that at 0<x<1 the area is approaching minus infinity whereas when x>1 the area is approaching infinity? I know that it's coming back to the same question of infinity minus infinity, but I still don't understand why. Also the transition from the expression in line 3 to line 4 seems incorrect to me.
            – JoisBack
            Sep 9 at 9:24











          • It suffices note that $int_1^inftyfraclog xx$ diverges to conclude that the whole integral diverges. Thiems topic was discussed many times here on MSE, I try to find some link to those discussions.
            – gimusi
            Sep 9 at 16:06







          • 1




            For the second issue I’ve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
            – gimusi
            Sep 9 at 16:08












          up vote
          6
          down vote



          accepted







          up vote
          6
          down vote



          accepted






          We have that



          $$int_0^inftyfracln(xpi^x)xdx
          =int_0^inftyfracln x +xln pi xdx
          =int_0^inftyfracln xxdx+int_0^inftyln pi ,dx$$



          and we can conclude here since both integrals diverge.



          Note that for the limit we have



          $$left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(1/x)-(1/x)lnpi$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(x)-(1/x)lnpi$$



          $$xlnpi -(1/x)lnpi to infty$$



          but it can't be used to evaluate the integral indeed for the improper integral we need to consider for $a>0$ the two integrals



          $$int_0^inftyfracln(xpi^x)xdx=int_0^a fracln(xpi^x)xdx+int_a^inftyfracln(xpi^x)xdx$$



          and the two limits at $0$ and $infty$ for the two integrals are independent and must be evaluated separetely.






          share|cite|improve this answer














          We have that



          $$int_0^inftyfracln(xpi^x)xdx
          =int_0^inftyfracln x +xln pi xdx
          =int_0^inftyfracln xxdx+int_0^inftyln pi ,dx$$



          and we can conclude here since both integrals diverge.



          Note that for the limit we have



          $$left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(1/x)-(1/x)lnpi$$



          $$frac12ln^2(x)+xlnpi - frac12ln^2(x)-(1/x)lnpi$$



          $$xlnpi -(1/x)lnpi to infty$$



          but it can't be used to evaluate the integral indeed for the improper integral we need to consider for $a>0$ the two integrals



          $$int_0^inftyfracln(xpi^x)xdx=int_0^a fracln(xpi^x)xdx+int_a^inftyfracln(xpi^x)xdx$$



          and the two limits at $0$ and $infty$ for the two integrals are independent and must be evaluated separetely.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 9 at 9:11

























          answered Sep 9 at 8:52









          gimusi

          74.3k73889




          74.3k73889











          • Big thanks to your effort for explaining. However, can you pls explain why integral of lnx/x from x =0 to x=inf diverges? I mean I can understand that from x=1 to inf, it diverges. But here there are both plus and minus infinity at . Visually, I plotted the graph of lnx/x on desmos and found that at 0<x<1 the area is approaching minus infinity whereas when x>1 the area is approaching infinity? I know that it's coming back to the same question of infinity minus infinity, but I still don't understand why. Also the transition from the expression in line 3 to line 4 seems incorrect to me.
            – JoisBack
            Sep 9 at 9:24











          • It suffices note that $int_1^inftyfraclog xx$ diverges to conclude that the whole integral diverges. Thiems topic was discussed many times here on MSE, I try to find some link to those discussions.
            – gimusi
            Sep 9 at 16:06







          • 1




            For the second issue I’ve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
            – gimusi
            Sep 9 at 16:08
















          • Big thanks to your effort for explaining. However, can you pls explain why integral of lnx/x from x =0 to x=inf diverges? I mean I can understand that from x=1 to inf, it diverges. But here there are both plus and minus infinity at . Visually, I plotted the graph of lnx/x on desmos and found that at 0<x<1 the area is approaching minus infinity whereas when x>1 the area is approaching infinity? I know that it's coming back to the same question of infinity minus infinity, but I still don't understand why. Also the transition from the expression in line 3 to line 4 seems incorrect to me.
            – JoisBack
            Sep 9 at 9:24











          • It suffices note that $int_1^inftyfraclog xx$ diverges to conclude that the whole integral diverges. Thiems topic was discussed many times here on MSE, I try to find some link to those discussions.
            – gimusi
            Sep 9 at 16:06







          • 1




            For the second issue I’ve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
            – gimusi
            Sep 9 at 16:08















          Big thanks to your effort for explaining. However, can you pls explain why integral of lnx/x from x =0 to x=inf diverges? I mean I can understand that from x=1 to inf, it diverges. But here there are both plus and minus infinity at . Visually, I plotted the graph of lnx/x on desmos and found that at 0<x<1 the area is approaching minus infinity whereas when x>1 the area is approaching infinity? I know that it's coming back to the same question of infinity minus infinity, but I still don't understand why. Also the transition from the expression in line 3 to line 4 seems incorrect to me.
          – JoisBack
          Sep 9 at 9:24





          Big thanks to your effort for explaining. However, can you pls explain why integral of lnx/x from x =0 to x=inf diverges? I mean I can understand that from x=1 to inf, it diverges. But here there are both plus and minus infinity at . Visually, I plotted the graph of lnx/x on desmos and found that at 0<x<1 the area is approaching minus infinity whereas when x>1 the area is approaching infinity? I know that it's coming back to the same question of infinity minus infinity, but I still don't understand why. Also the transition from the expression in line 3 to line 4 seems incorrect to me.
          – JoisBack
          Sep 9 at 9:24













          It suffices note that $int_1^inftyfraclog xx$ diverges to conclude that the whole integral diverges. Thiems topic was discussed many times here on MSE, I try to find some link to those discussions.
          – gimusi
          Sep 9 at 16:06





          It suffices note that $int_1^inftyfraclog xx$ diverges to conclude that the whole integral diverges. Thiems topic was discussed many times here on MSE, I try to find some link to those discussions.
          – gimusi
          Sep 9 at 16:06





          1




          1




          For the second issue I’ve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
          – gimusi
          Sep 9 at 16:08




          For the second issue I’ve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
          – gimusi
          Sep 9 at 16:08

















           

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