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
$begingroup$
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.
calculus integration
asked Mar 14 at 6:32
Edward Z. MiaoEdward Z. Miao
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.
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add a comment |
$begingroup$
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.
calculus integration
asked Mar 14 at 6:32
Edward Z. MiaoEdward Z. Miao
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
calculus integration
asked Mar 14 at 6:32
Edward Z. MiaoEdward Z. Miao
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.
$endgroup$
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
calculus integration
asked Mar 14 at 6:32
Edward Z. MiaoEdward Z. Miao
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.
asked Mar 14 at 6:32
Edward Z. MiaoEdward Z. Miao
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.
asked Mar 14 at 6:32
Edward Z. MiaoEdward Z. Miao
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.
asked Mar 14 at 6:32
Edward Z. MiaoEdward Z. Miao
113
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $a<-2$.
answered Mar 14 at 6:57
J.G.J.G.
30.9k23149
$endgroup$
add a comment |
$begingroup$
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 $aleq 2$ )
answered Mar 14 at 8:53
FDPFDP
6,17211829
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $a<-2$.
answered Mar 14 at 6:57
J.G.J.G.
30.9k23149
$endgroup$
add a comment |
$begingroup$
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 $a<-2$.
answered Mar 14 at 6:57
J.G.J.G.
30.9k23149
$endgroup$
add a comment |
$begingroup$
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 $a<-2$.
answered Mar 14 at 6:57
J.G.J.G.
30.9k23149
$endgroup$
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 $a<-2$.
answered Mar 14 at 6:57
J.G.J.G.
30.9k23149
edited Mar 14 at 7:12
answered Mar 14 at 6:57
J.G.J.G.
30.9k23149
answered Mar 14 at 6:57
J.G.J.G.
30.9k23149
answered Mar 14 at 6:57
J.G.J.G.
30.9k23149
30.9k23149
add a comment |
add a comment |
$begingroup$
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 $aleq 2$ )
answered Mar 14 at 8:53
FDPFDP
6,17211829
$endgroup$
add a comment |
$begingroup$
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 $aleq 2$ )
answered Mar 14 at 8:53
FDPFDP
6,17211829
$endgroup$
add a comment |
$begingroup$
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 $aleq 2$ )
answered Mar 14 at 8:53
FDPFDP
6,17211829
$endgroup$
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 $aleq 2$ )
answered Mar 14 at 8:53
FDPFDP
6,17211829
answered Mar 14 at 8:53
FDPFDP
6,17211829
answered Mar 14 at 8:53
FDPFDP
6,17211829
answered Mar 14 at 8:53
FDPFDP
6,17211829
6,17211829
add a comment |
add a comment |
Edward Z. Miao is a new contributor. Be nice, and check out our Code of Conduct.
Edward Z. Miao is a new contributor. Be nice, and check out our Code of Conduct.
Edward Z. Miao is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
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