Two answers for Integral of $ int_0^infty fractan^-1ax - tan^-1xxmathrm dx$Two integrals on the form $int_0^infty x f(x^3) mathrmdx$about the gaussian integral $int_0^infty e^-x^2 mathrmdx$How to find $int_0^infty prod_k=1^n fracsin fracx2k-1fracx2k-1mathrm dx$Integral $int_0^pi/2 ln(1+alphasin^2 x), dx=pi ln frac1+sqrt1+alpha2$Evaluating $int_0^2(tan^-1(pi x)-tan^-1 x),mathrmdx$$int_-infty^inftye^x+1over (e^x-x+1)^2+pi^2mathrm dx=int_-infty^inftye^x+1over (e^x+x+1)^2+pi^2mathrm dx=1$Solution of $int_0^infty fracarctan(x)-arctan(2x)xmathrm dx$ : Why Wolfram Alpha shows a different answer than mine?Evaluate $int_0^inftyfractan xx^ndx$Integrate $int_0^inftyfrac(1+x^2)dx(a^2+b^2x^2)^2$Integral $int_0^fracpi2 x^2sqrttan x,mathrm dx$
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Two answers for Integral of $ int_0^infty fractan^-1ax - tan^-1xxmathrm dx$
Two integrals on the form $int_0^infty x f(x^3) mathrmdx$about the gaussian integral $int_0^infty e^-x^2 mathrmdx$How to find $int_0^infty prod_k=1^n fracsin fracx2k-1fracx2k-1mathrm dx$Integral $int_0^pi/2 ln(1+alphasin^2 x), dx=pi ln frac1+sqrt1+alpha2$Evaluating $int_0^2(tan^-1(pi x)-tan^-1 x),mathrmdx$$int_-infty^inftye^x+1over (e^x-x+1)^2+pi^2mathrm dx=int_-infty^inftye^x+1over (e^x+x+1)^2+pi^2mathrm dx=1$Solution of $int_0^infty fracarctan(x)-arctan(2x)xmathrm dx$ : Why Wolfram Alpha shows a different answer than mine?Evaluate $int_0^inftyfractan xx^ndx$Integrate $int_0^inftyfrac(1+x^2)dx(a^2+b^2x^2)^2$Integral $int_0^fracpi2 x^2sqrttan x,mathrm dx$
$begingroup$
$$ I= int_0^infty fractan^-1ax - tan^-1xxmathrm dx$$
Now by using Leibnitz Rule by differentiating w.r.t. to $a$ we get,
$$fracmathrm dImathrm da=fracpi2a$$
$$I=fracpi2ln a$$
But consider ,
$$I_1= int_0^infty fractan^-1axxmathrm dx$$
$$I_2= int_0^infty fractan^-1xxmathrm dx$$
So
Substituting $ax=t$ in $I_1$
$$I_1= int_0^infty fractan^-1ttmathrm dt$$
So $$I=I_1-I_2=0$$
So where am I wrong here? I know there is some mistake
In second method as the function is always positive integral can't be $0$. But i am not able to figure out where am I wrong.
Note: $a$ is a positive number
integration definite-integrals
edited Mar 13 at 8:51
mrtaurho
6,00551641
asked Mar 13 at 8:47
ATHARVAATHARVA
476312
$endgroup$
add a comment |
$begingroup$
$$ I= int_0^infty fractan^-1ax - tan^-1xxmathrm dx$$
Now by using Leibnitz Rule by differentiating w.r.t. to $a$ we get,
$$fracmathrm dImathrm da=fracpi2a$$
$$I=fracpi2ln a$$
But consider ,
$$I_1= int_0^infty fractan^-1axxmathrm dx$$
$$I_2= int_0^infty fractan^-1xxmathrm dx$$
So
Substituting $ax=t$ in $I_1$
$$I_1= int_0^infty fractan^-1ttmathrm dt$$
So $$I=I_1-I_2=0$$
So where am I wrong here? I know there is some mistake
In second method as the function is always positive integral can't be $0$. But i am not able to figure out where am I wrong.
Note: $a$ is a positive number
integration definite-integrals
edited Mar 13 at 8:51
mrtaurho
6,00551641
asked Mar 13 at 8:47
ATHARVAATHARVA
476312
$endgroup$
1
$begingroup$
$I_1$ diverges at its upper limit due to a $O(1/x)$ integrand. Incidentally, your first, correct strategy is common.
$endgroup$
– J.G.
Mar 15 at 12:31
add a comment |
$begingroup$
$$ I= int_0^infty fractan^-1ax - tan^-1xxmathrm dx$$
Now by using Leibnitz Rule by differentiating w.r.t. to $a$ we get,
$$fracmathrm dImathrm da=fracpi2a$$
$$I=fracpi2ln a$$
But consider ,
$$I_1= int_0^infty fractan^-1axxmathrm dx$$
$$I_2= int_0^infty fractan^-1xxmathrm dx$$
So
Substituting $ax=t$ in $I_1$
$$I_1= int_0^infty fractan^-1ttmathrm dt$$
So $$I=I_1-I_2=0$$
So where am I wrong here? I know there is some mistake
In second method as the function is always positive integral can't be $0$. But i am not able to figure out where am I wrong.
Note: $a$ is a positive number
integration definite-integrals
edited Mar 13 at 8:51
mrtaurho
6,00551641
asked Mar 13 at 8:47
ATHARVAATHARVA
476312
$endgroup$
$$ I= int_0^infty fractan^-1ax - tan^-1xxmathrm dx$$
Now by using Leibnitz Rule by differentiating w.r.t. to $a$ we get,
$$fracmathrm dImathrm da=fracpi2a$$
$$I=fracpi2ln a$$
But consider ,
$$I_1= int_0^infty fractan^-1axxmathrm dx$$
$$I_2= int_0^infty fractan^-1xxmathrm dx$$
So
Substituting $ax=t$ in $I_1$
$$I_1= int_0^infty fractan^-1ttmathrm dt$$
So $$I=I_1-I_2=0$$
So where am I wrong here? I know there is some mistake
In second method as the function is always positive integral can't be $0$. But i am not able to figure out where am I wrong.
Note: $a$ is a positive number
integration definite-integrals
integration definite-integrals
edited Mar 13 at 8:51
mrtaurho
6,00551641
asked Mar 13 at 8:47
ATHARVAATHARVA
476312
edited Mar 13 at 8:51
mrtaurho
6,00551641
asked Mar 13 at 8:47
ATHARVAATHARVA
476312
edited Mar 13 at 8:51
mrtaurho
6,00551641
edited Mar 13 at 8:51
mrtaurho
6,00551641
edited Mar 13 at 8:51
mrtaurho
6,00551641
6,00551641
asked Mar 13 at 8:47
ATHARVAATHARVA
476312
asked Mar 13 at 8:47
ATHARVAATHARVA
476312
asked Mar 13 at 8:47
ATHARVAATHARVA
476312
476312
1
$begingroup$
$I_1$ diverges at its upper limit due to a $O(1/x)$ integrand. Incidentally, your first, correct strategy is common.
$endgroup$
– J.G.
Mar 15 at 12:31
add a comment |
1
$begingroup$
$I_1$ diverges at its upper limit due to a $O(1/x)$ integrand. Incidentally, your first, correct strategy is common.
$endgroup$
– J.G.
Mar 15 at 12:31
1
1
$begingroup$
$I_1$ diverges at its upper limit due to a $O(1/x)$ integrand. Incidentally, your first, correct strategy is common.
$endgroup$
– J.G.
Mar 15 at 12:31
$begingroup$
$I_1$ diverges at its upper limit due to a $O(1/x)$ integrand. Incidentally, your first, correct strategy is common.
$endgroup$
– J.G.
Mar 15 at 12:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are wrong when you split the original integral into two integrals. This can be done only when both the splitted integrals converge to some finite value, which is not the case here. So your second method fails. The correct answer hence is $$I(a)=frac pi2ln a$$
answered Mar 13 at 8:54
DarkraiDarkrai
6,4371442
$endgroup$
3
$begingroup$
In other words, the integrals $I_1, I_2$ both diverge. This is one of many examples where divergent integrals may not be treated with the rules for convergent integrals.
$endgroup$
– GEdgar
Mar 13 at 10:42
add a comment |
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$begingroup$
You are wrong when you split the original integral into two integrals. This can be done only when both the splitted integrals converge to some finite value, which is not the case here. So your second method fails. The correct answer hence is $$I(a)=frac pi2ln a$$
answered Mar 13 at 8:54
DarkraiDarkrai
6,4371442
$endgroup$
3
$begingroup$
In other words, the integrals $I_1, I_2$ both diverge. This is one of many examples where divergent integrals may not be treated with the rules for convergent integrals.
$endgroup$
– GEdgar
Mar 13 at 10:42
add a comment |
$begingroup$
You are wrong when you split the original integral into two integrals. This can be done only when both the splitted integrals converge to some finite value, which is not the case here. So your second method fails. The correct answer hence is $$I(a)=frac pi2ln a$$
answered Mar 13 at 8:54
DarkraiDarkrai
6,4371442
$endgroup$
3
$begingroup$
In other words, the integrals $I_1, I_2$ both diverge. This is one of many examples where divergent integrals may not be treated with the rules for convergent integrals.
$endgroup$
– GEdgar
Mar 13 at 10:42
add a comment |
$begingroup$
You are wrong when you split the original integral into two integrals. This can be done only when both the splitted integrals converge to some finite value, which is not the case here. So your second method fails. The correct answer hence is $$I(a)=frac pi2ln a$$
answered Mar 13 at 8:54
DarkraiDarkrai
6,4371442
$endgroup$
You are wrong when you split the original integral into two integrals. This can be done only when both the splitted integrals converge to some finite value, which is not the case here. So your second method fails. The correct answer hence is $$I(a)=frac pi2ln a$$
answered Mar 13 at 8:54
DarkraiDarkrai
6,4371442
answered Mar 13 at 8:54
DarkraiDarkrai
6,4371442
answered Mar 13 at 8:54
DarkraiDarkrai
6,4371442
answered Mar 13 at 8:54
DarkraiDarkrai
6,4371442
6,4371442
3
$begingroup$
In other words, the integrals $I_1, I_2$ both diverge. This is one of many examples where divergent integrals may not be treated with the rules for convergent integrals.
$endgroup$
– GEdgar
Mar 13 at 10:42
add a comment |
3
$begingroup$
In other words, the integrals $I_1, I_2$ both diverge. This is one of many examples where divergent integrals may not be treated with the rules for convergent integrals.
$endgroup$
– GEdgar
Mar 13 at 10:42
3
3
$begingroup$
In other words, the integrals $I_1, I_2$ both diverge. This is one of many examples where divergent integrals may not be treated with the rules for convergent integrals.
$endgroup$
– GEdgar
Mar 13 at 10:42
$begingroup$
In other words, the integrals $I_1, I_2$ both diverge. This is one of many examples where divergent integrals may not be treated with the rules for convergent integrals.
$endgroup$
– GEdgar
Mar 13 at 10:42
add a comment |
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$begingroup$
$I_1$ diverges at its upper limit due to a $O(1/x)$ integrand. Incidentally, your first, correct strategy is common.
$endgroup$
– J.G.
Mar 15 at 12:31