Find the integral of log(1 - a/x^2 + 1/x^4) from 0 to infinityProving $int_0^infty logleft (1-2fraccos 2thetax^2+frac1x^4 right)dx =2pi sin theta$Convergence of Fourier integralImproper Multiple IntegralIntegral form for the euler-mascheroni gamma constant using floor functionHow to prove $operatornameLog(z) = log(|z|)+iarg(z)$.Prove the given two integrals are not equalNecessary condition for the convergence of an improper integral.Is $int_0^afracf(x)^2int_0^x f(t),dt,dxgeqint_0^a fracg(x)^2int_0^xg(t),dt,dx.$?Evaluate $int_0^inftyfraclog^2 xe^x^2mathrmdx$.What's about functions $f:[0,infty)tomathbbR$ satisfying $int_0^infty f(x)(log x)dx=1$?Test $int_0^inftyfrace^ixlog(x), int_0^inftyfraccos(x)x^p, int_0^inftycos(x^2)$ for convergence

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Find the integral of log(1 - a/x^2 + 1/x^4) from 0 to infinity


Proving $int_0^infty logleft (1-2fraccos 2thetax^2+frac1x^4 right)dx =2pi sin theta$Convergence of Fourier integralImproper Multiple IntegralIntegral form for the euler-mascheroni gamma constant using floor functionHow to prove $operatornameLog(z) = log(|z|)+iarg(z)$.Prove the given two integrals are not equalNecessary condition for the convergence of an improper integral.Is $int_0^afracf(x)^2int_0^x f(t),dt,dxgeqint_0^a fracg(x)^2int_0^xg(t),dt,dx.$?Evaluate $int_0^inftyfraclog^2 xe^x^2mathrmdx$.What's about functions $f:[0,infty)tomathbbR$ satisfying $int_0^infty f(x)(log x)dx=1$?Test $int_0^inftyfrace^ixlog(x), int_0^inftyfraccos(x)x^p, int_0^inftycos(x^2)$ for convergence













2












$begingroup$


It is a exercise from Calculus appeared in improper integral, one of my classmates ask me. It reads as follows:
$int_0^infty log(1-fracax^2+frac1x^4) dx = ?$.


I have tried to let t = $frac1x$ and rewrite LHS as $int_0^infty fraclog(1-fracax^2+frac1x^4)x^2 dx$. Or write it as $int_0^infty log(x^4 - ax^2 +1) dx$ $-int_0^infty 4log(x) dx$ in a more simple form. But I still cannot calculate it directly. I wonder whether we should use more delicate techniques to solve the problem?

I know it does not contain any philosophy in this problem but I would like to regard it as an intelligent challenge.

I would appreciate it if anyone can give me some insights.










share|cite|improve this question







New contributor




Edward Z. Miao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$











  • $begingroup$
    What is $a$. For large values of $a$ the integrand is not well defined.
    $endgroup$
    – Kavi Rama Murthy
    Mar 14 at 6:45










  • $begingroup$
    $int_0^infty log(x)dx$ doesn't exist.
    $endgroup$
    – FDP
    Mar 14 at 8:43















2












$begingroup$


It is a exercise from Calculus appeared in improper integral, one of my classmates ask me. It reads as follows:
$int_0^infty log(1-fracax^2+frac1x^4) dx = ?$.


I have tried to let t = $frac1x$ and rewrite LHS as $int_0^infty fraclog(1-fracax^2+frac1x^4)x^2 dx$. Or write it as $int_0^infty log(x^4 - ax^2 +1) dx$ $-int_0^infty 4log(x) dx$ in a more simple form. But I still cannot calculate it directly. I wonder whether we should use more delicate techniques to solve the problem?

I know it does not contain any philosophy in this problem but I would like to regard it as an intelligent challenge.

I would appreciate it if anyone can give me some insights.










share|cite|improve this question







New contributor




Edward Z. Miao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    What is $a$. For large values of $a$ the integrand is not well defined.
    $endgroup$
    – Kavi Rama Murthy
    Mar 14 at 6:45










  • $begingroup$
    $int_0^infty log(x)dx$ doesn't exist.
    $endgroup$
    – FDP
    Mar 14 at 8:43













2












2








2


0



$begingroup$


It is a exercise from Calculus appeared in improper integral, one of my classmates ask me. It reads as follows:
$int_0^infty log(1-fracax^2+frac1x^4) dx = ?$.


I have tried to let t = $frac1x$ and rewrite LHS as $int_0^infty fraclog(1-fracax^2+frac1x^4)x^2 dx$. Or write it as $int_0^infty log(x^4 - ax^2 +1) dx$ $-int_0^infty 4log(x) dx$ in a more simple form. But I still cannot calculate it directly. I wonder whether we should use more delicate techniques to solve the problem?

I know it does not contain any philosophy in this problem but I would like to regard it as an intelligent challenge.

I would appreciate it if anyone can give me some insights.










share|cite|improve this question







New contributor




Edward Z. Miao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




It is a exercise from Calculus appeared in improper integral, one of my classmates ask me. It reads as follows:
$int_0^infty log(1-fracax^2+frac1x^4) dx = ?$.


I have tried to let t = $frac1x$ and rewrite LHS as $int_0^infty fraclog(1-fracax^2+frac1x^4)x^2 dx$. Or write it as $int_0^infty log(x^4 - ax^2 +1) dx$ $-int_0^infty 4log(x) dx$ in a more simple form. But I still cannot calculate it directly. I wonder whether we should use more delicate techniques to solve the problem?

I know it does not contain any philosophy in this problem but I would like to regard it as an intelligent challenge.

I would appreciate it if anyone can give me some insights.







calculus integration






share|cite|improve this question







New contributor




Edward Z. Miao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Edward Z. Miao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Edward Z. Miao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Mar 14 at 6:32









Edward Z. MiaoEdward Z. Miao

113




113




New contributor




Edward Z. Miao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Edward Z. Miao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Edward Z. Miao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    What is $a$. For large values of $a$ the integrand is not well defined.
    $endgroup$
    – Kavi Rama Murthy
    Mar 14 at 6:45










  • $begingroup$
    $int_0^infty log(x)dx$ doesn't exist.
    $endgroup$
    – FDP
    Mar 14 at 8:43
















  • $begingroup$
    What is $a$. For large values of $a$ the integrand is not well defined.
    $endgroup$
    – Kavi Rama Murthy
    Mar 14 at 6:45










  • $begingroup$
    $int_0^infty log(x)dx$ doesn't exist.
    $endgroup$
    – FDP
    Mar 14 at 8:43















$begingroup$
What is $a$. For large values of $a$ the integrand is not well defined.
$endgroup$
– Kavi Rama Murthy
Mar 14 at 6:45




$begingroup$
What is $a$. For large values of $a$ the integrand is not well defined.
$endgroup$
– Kavi Rama Murthy
Mar 14 at 6:45












$begingroup$
$int_0^infty log(x)dx$ doesn't exist.
$endgroup$
– FDP
Mar 14 at 8:43




$begingroup$
$int_0^infty log(x)dx$ doesn't exist.
$endgroup$
– FDP
Mar 14 at 8:43










2 Answers
2






active

oldest

votes


















2












$begingroup$

If $|a|le 2$, the parameterisation $a=2cos 2theta$ obtains the answer as $2pisintheta$, i.e. $pisqrt2-a$. If $a>2$, there are real values of $a$ for which the logarithm's argument is negative, so that the integrand is non-real. The former case can presumably be analytically continued to $a<-2$.






share|cite|improve this answer











$endgroup$




















    1












    $begingroup$

    See my answer here: https://math.stackexchange.com/a/3144140/186817



    (replace $a$ by $-fraca2$ into the formula)



    The desired result is $pisqrt2-a$



    (Observe that one needs $aleq 2$ )






    share|cite|improve this answer









    $endgroup$












      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      If $|a|le 2$, the parameterisation $a=2cos 2theta$ obtains the answer as $2pisintheta$, i.e. $pisqrt2-a$. If $a>2$, there are real values of $a$ for which the logarithm's argument is negative, so that the integrand is non-real. The former case can presumably be analytically continued to $a<-2$.






      share|cite|improve this answer











      $endgroup$

















        2












        $begingroup$

        If $|a|le 2$, the parameterisation $a=2cos 2theta$ obtains the answer as $2pisintheta$, i.e. $pisqrt2-a$. If $a>2$, there are real values of $a$ for which the logarithm's argument is negative, so that the integrand is non-real. The former case can presumably be analytically continued to $a<-2$.






        share|cite|improve this answer











        $endgroup$















          2












          2








          2





          $begingroup$

          If $|a|le 2$, the parameterisation $a=2cos 2theta$ obtains the answer as $2pisintheta$, i.e. $pisqrt2-a$. If $a>2$, there are real values of $a$ for which the logarithm's argument is negative, so that the integrand is non-real. The former case can presumably be analytically continued to $a<-2$.






          share|cite|improve this answer











          $endgroup$



          If $|a|le 2$, the parameterisation $a=2cos 2theta$ obtains the answer as $2pisintheta$, i.e. $pisqrt2-a$. If $a>2$, there are real values of $a$ for which the logarithm's argument is negative, so that the integrand is non-real. The former case can presumably be analytically continued to $a<-2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 14 at 7:12

























          answered Mar 14 at 6:57









          J.G.J.G.

          30.9k23149




          30.9k23149





















              1












              $begingroup$

              See my answer here: https://math.stackexchange.com/a/3144140/186817



              (replace $a$ by $-fraca2$ into the formula)



              The desired result is $pisqrt2-a$



              (Observe that one needs $aleq 2$ )






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                See my answer here: https://math.stackexchange.com/a/3144140/186817



                (replace $a$ by $-fraca2$ into the formula)



                The desired result is $pisqrt2-a$



                (Observe that one needs $aleq 2$ )






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  See my answer here: https://math.stackexchange.com/a/3144140/186817



                  (replace $a$ by $-fraca2$ into the formula)



                  The desired result is $pisqrt2-a$



                  (Observe that one needs $aleq 2$ )






                  share|cite|improve this answer









                  $endgroup$



                  See my answer here: https://math.stackexchange.com/a/3144140/186817



                  (replace $a$ by $-fraca2$ into the formula)



                  The desired result is $pisqrt2-a$



                  (Observe that one needs $aleq 2$ )







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 14 at 8:53









                  FDPFDP

                  6,17211829




                  6,17211829




















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