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$int_0^1fraclnxln(1+x)1+xdx$
The Next CEO of Stack OverflowWays to prove $ int_0^1 fracln^2(1+x)xdx = fraczeta(3)4$?how to evaluate $int_0^fracpi2frac1sqrtsin xtextdx$Integral $I:=int_0^1 fraclog^2 xx^2-x+1mathrm dx=frac10pi^381 sqrt 3}$Integral $int_0^infty frac{sqrt[3]x+1 - sqrt[3]xsqrtx , mathrm dx$Integral of Bessel function multiplied with sine $int_0^infty J_0(bx) sin(ax) dx$.Definite integral problem of $fracx^nn!$Derivative of improper Integral $f(t)= int_0^1 fracsin(xt)x:dx$Ways to evaluate $int_0^1 int_0^1 frac11-xydxdy = fracpi^26$Asymptotics of a double integral: $ int_0^inftyduint_0^inftydv, frac1(u+v)^2expleft(-fracxu+vright)$Calculate an approximation of $int_0^1int_0^1fraclog(xy)xy-1+log(xy)dxdy$Show that $int_0^inftyfracoperatornameLi_s(-x)x^alpha+1mathrm dx=-frac1alpha^sfracpisin(pi alpha)$
$begingroup$
I want to solve for the following Integral:
$$int_0^1fraclnxln(1+x)1+xdx$$
I have tried to use:
$$ln(1+x)=-sum_k=1^inftyfrac(-1)^kx^kk$$
and so
$$int_0^1fraclnxln(1+x)1+xdx=-sum_k=1^inftyfrac(-1)^kkint_0^1fracx^klnx1+xdx$$
definite-integrals
edited Mar 28 at 10:16
Milten
3226
asked Mar 28 at 9:59
Reynan HenryReynan Henry
801
$endgroup$
add a comment |
$begingroup$
I want to solve for the following Integral:
$$int_0^1fraclnxln(1+x)1+xdx$$
I have tried to use:
$$ln(1+x)=-sum_k=1^inftyfrac(-1)^kx^kk$$
and so
$$int_0^1fraclnxln(1+x)1+xdx=-sum_k=1^inftyfrac(-1)^kkint_0^1fracx^klnx1+xdx$$
definite-integrals
edited Mar 28 at 10:16
Milten
3226
asked Mar 28 at 9:59
Reynan HenryReynan Henry
801
$endgroup$
add a comment |
$begingroup$
I want to solve for the following Integral:
$$int_0^1fraclnxln(1+x)1+xdx$$
I have tried to use:
$$ln(1+x)=-sum_k=1^inftyfrac(-1)^kx^kk$$
and so
$$int_0^1fraclnxln(1+x)1+xdx=-sum_k=1^inftyfrac(-1)^kkint_0^1fracx^klnx1+xdx$$
definite-integrals
edited Mar 28 at 10:16
Milten
3226
asked Mar 28 at 9:59
Reynan HenryReynan Henry
801
$endgroup$
I want to solve for the following Integral:
$$int_0^1fraclnxln(1+x)1+xdx$$
I have tried to use:
$$ln(1+x)=-sum_k=1^inftyfrac(-1)^kx^kk$$
and so
$$int_0^1fraclnxln(1+x)1+xdx=-sum_k=1^inftyfrac(-1)^kkint_0^1fracx^klnx1+xdx$$
definite-integrals
definite-integrals
edited Mar 28 at 10:16
Milten
3226
asked Mar 28 at 9:59
Reynan HenryReynan Henry
801
edited Mar 28 at 10:16
Milten
3226
asked Mar 28 at 9:59
Reynan HenryReynan Henry
801
edited Mar 28 at 10:16
Milten
3226
edited Mar 28 at 10:16
Milten
3226
edited Mar 28 at 10:16
Milten
3226
3226
asked Mar 28 at 9:59
Reynan HenryReynan Henry
801
asked Mar 28 at 9:59
Reynan HenryReynan Henry
801
asked Mar 28 at 9:59
Reynan HenryReynan Henry
801
801
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: without power series: use the substitution $t=ln(1+x)$.
answered Mar 28 at 10:04
FredFred
48.7k11849
$endgroup$
add a comment |
$begingroup$
The integral is $frac 1 2int_0^1 f(g^2)'(x)dx$ where $g(x)=log (1+x)$. Integrating by parts we get $frac 1 2 [fg^2|_0^1-int_0^1 frac g(x)^2 x dx]$. To compute the integral in the second term make the substitution $y=log (1+x)$. You will now get something familiar and I will let you complete the evaluation.
answered Mar 28 at 10:07
Kavi Rama MurthyKavi Rama Murthy
71.6k53170
$endgroup$
add a comment |
$begingroup$
Let
$$I = int_0^1 fracln x ln (1 + x)1 + x , dx.$$
Integrating by parts we have
$$I = - frac12 int_0^1 fracln^2 (1 + x)x , dx tag1$$
The integral appearing in (1) can be calculated in various ways. One way I have already shown here. The result is $zeta (3)/4$ where $zeta (z)$ is the Riemann zeta function. Thus
$$int_0^1 fracln x ln (1 + x)1 + x , dx = -frac18 zeta (3).$$
answered Mar 28 at 10:16
omegadotomegadot
6,2592829
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: without power series: use the substitution $t=ln(1+x)$.
answered Mar 28 at 10:04
FredFred
48.7k11849
$endgroup$
add a comment |
$begingroup$
Hint: without power series: use the substitution $t=ln(1+x)$.
answered Mar 28 at 10:04
FredFred
48.7k11849
$endgroup$
add a comment |
$begingroup$
Hint: without power series: use the substitution $t=ln(1+x)$.
answered Mar 28 at 10:04
FredFred
48.7k11849
$endgroup$
Hint: without power series: use the substitution $t=ln(1+x)$.
answered Mar 28 at 10:04
FredFred
48.7k11849
answered Mar 28 at 10:04
FredFred
48.7k11849
answered Mar 28 at 10:04
FredFred
48.7k11849
answered Mar 28 at 10:04
FredFred
48.7k11849
48.7k11849
add a comment |
add a comment |
$begingroup$
The integral is $frac 1 2int_0^1 f(g^2)'(x)dx$ where $g(x)=log (1+x)$. Integrating by parts we get $frac 1 2 [fg^2|_0^1-int_0^1 frac g(x)^2 x dx]$. To compute the integral in the second term make the substitution $y=log (1+x)$. You will now get something familiar and I will let you complete the evaluation.
answered Mar 28 at 10:07
Kavi Rama MurthyKavi Rama Murthy
71.6k53170
$endgroup$
add a comment |
$begingroup$
The integral is $frac 1 2int_0^1 f(g^2)'(x)dx$ where $g(x)=log (1+x)$. Integrating by parts we get $frac 1 2 [fg^2|_0^1-int_0^1 frac g(x)^2 x dx]$. To compute the integral in the second term make the substitution $y=log (1+x)$. You will now get something familiar and I will let you complete the evaluation.
answered Mar 28 at 10:07
Kavi Rama MurthyKavi Rama Murthy
71.6k53170
$endgroup$
add a comment |
$begingroup$
The integral is $frac 1 2int_0^1 f(g^2)'(x)dx$ where $g(x)=log (1+x)$. Integrating by parts we get $frac 1 2 [fg^2|_0^1-int_0^1 frac g(x)^2 x dx]$. To compute the integral in the second term make the substitution $y=log (1+x)$. You will now get something familiar and I will let you complete the evaluation.
answered Mar 28 at 10:07
Kavi Rama MurthyKavi Rama Murthy
71.6k53170
$endgroup$
The integral is $frac 1 2int_0^1 f(g^2)'(x)dx$ where $g(x)=log (1+x)$. Integrating by parts we get $frac 1 2 [fg^2|_0^1-int_0^1 frac g(x)^2 x dx]$. To compute the integral in the second term make the substitution $y=log (1+x)$. You will now get something familiar and I will let you complete the evaluation.
answered Mar 28 at 10:07
Kavi Rama MurthyKavi Rama Murthy
71.6k53170
answered Mar 28 at 10:07
Kavi Rama MurthyKavi Rama Murthy
71.6k53170
answered Mar 28 at 10:07
Kavi Rama MurthyKavi Rama Murthy
71.6k53170
answered Mar 28 at 10:07
Kavi Rama MurthyKavi Rama Murthy
71.6k53170
71.6k53170
add a comment |
add a comment |
$begingroup$
Let
$$I = int_0^1 fracln x ln (1 + x)1 + x , dx.$$
Integrating by parts we have
$$I = - frac12 int_0^1 fracln^2 (1 + x)x , dx tag1$$
The integral appearing in (1) can be calculated in various ways. One way I have already shown here. The result is $zeta (3)/4$ where $zeta (z)$ is the Riemann zeta function. Thus
$$int_0^1 fracln x ln (1 + x)1 + x , dx = -frac18 zeta (3).$$
answered Mar 28 at 10:16
omegadotomegadot
6,2592829
$endgroup$
add a comment |
$begingroup$
Let
$$I = int_0^1 fracln x ln (1 + x)1 + x , dx.$$
Integrating by parts we have
$$I = - frac12 int_0^1 fracln^2 (1 + x)x , dx tag1$$
The integral appearing in (1) can be calculated in various ways. One way I have already shown here. The result is $zeta (3)/4$ where $zeta (z)$ is the Riemann zeta function. Thus
$$int_0^1 fracln x ln (1 + x)1 + x , dx = -frac18 zeta (3).$$
answered Mar 28 at 10:16
omegadotomegadot
6,2592829
$endgroup$
add a comment |
$begingroup$
Let
$$I = int_0^1 fracln x ln (1 + x)1 + x , dx.$$
Integrating by parts we have
$$I = - frac12 int_0^1 fracln^2 (1 + x)x , dx tag1$$
The integral appearing in (1) can be calculated in various ways. One way I have already shown here. The result is $zeta (3)/4$ where $zeta (z)$ is the Riemann zeta function. Thus
$$int_0^1 fracln x ln (1 + x)1 + x , dx = -frac18 zeta (3).$$
answered Mar 28 at 10:16
omegadotomegadot
6,2592829
$endgroup$
Let
$$I = int_0^1 fracln x ln (1 + x)1 + x , dx.$$
Integrating by parts we have
$$I = - frac12 int_0^1 fracln^2 (1 + x)x , dx tag1$$
The integral appearing in (1) can be calculated in various ways. One way I have already shown here. The result is $zeta (3)/4$ where $zeta (z)$ is the Riemann zeta function. Thus
$$int_0^1 fracln x ln (1 + x)1 + x , dx = -frac18 zeta (3).$$
answered Mar 28 at 10:16
omegadotomegadot
6,2592829
answered Mar 28 at 10:16
omegadotomegadot
6,2592829
answered Mar 28 at 10:16
omegadotomegadot
6,2592829
answered Mar 28 at 10:16
omegadotomegadot
6,2592829
6,2592829
add a comment |
add a comment |
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