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:
- Does A=B (doesn't seem so)
- Does the integral converges or diverges? (And how to figure that out)
calculus improper-integrals
add a comment |Â
up vote
4
down vote
favorite
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:
- Does A=B (doesn't seem so)
- Does the integral converges or diverges? (And how to figure that out)
calculus improper-integrals
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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:
- Does A=B (doesn't seem so)
- Does the integral converges or diverges? (And how to figure that out)
calculus improper-integrals
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:
- Does A=B (doesn't seem so)
- Does the integral converges or diverges? (And how to figure that out)
calculus improper-integrals
calculus improper-integrals
edited Sep 9 at 9:02
Bernard
112k636105
112k636105
asked Sep 9 at 8:48
JoisBack
1355
1355
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1 Answer
1
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up vote
6
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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.
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.
â 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.
â gimusi
Sep 9 at 16:06
1
For the second issue IâÂÂve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
â gimusi
Sep 9 at 16:08
add a comment |Â
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.
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.
â 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.
â gimusi
Sep 9 at 16:06
1
For the second issue IâÂÂve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
â gimusi
Sep 9 at 16:08
add a comment |Â
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.
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.
â 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.
â gimusi
Sep 9 at 16:06
1
For the second issue IâÂÂve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
â gimusi
Sep 9 at 16:08
add a comment |Â
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.
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.
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.
â 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.
â gimusi
Sep 9 at 16:06
1
For the second issue IâÂÂve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
â gimusi
Sep 9 at 16:08
add a comment |Â
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.
â 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.
â gimusi
Sep 9 at 16:06
1
For the second issue IâÂÂve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
â 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.
â 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.
â 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.
â 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.
â gimusi
Sep 9 at 16:06
1
1
For the second issue IâÂÂve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
â gimusi
Sep 9 at 16:08
For the second issue IâÂÂve used that $$log frac1x=-log ximplies log^2 frac1x=(-log x)^2=log^2 x$$
â gimusi
Sep 9 at 16:08
add a comment |Â
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