Discussing the convergence of $int_-1^1 dx/x^4$ The 2019 Stack Overflow Developer Survey Results Are InImproper integral $int^pi/2_0 (operatornamecsc x - frac1x),mathrm dx$Improper integral:$ int_-1^2 frac1xdx$ diverges$int_-infty^infty x^2 cdot frac12Ce^-C,dx$ where C is a constantConvergence of $int_-1^1fracx-1x^5/3dx$Convergence of $int_0^1 fracsqrt e^2+x^2 - e^cos xtan^axdx$Differing (divergent) limits in the improper integral $int_-1^3x^-3dx$Calculate $int_-infty^+inftyfracx1+x^2dx$, what is wrong with this?$int_-1^0 frac(e^1/x)x^2 dx$ Improper integralsEvaluating $int_-infty^infty 1-e^-frac1x^2rm dx$The convergence of a strange integral: $int_0^1fracx-1ln xmathbb dx$
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Discussing the convergence of $int_-1^1 dx/x^4$
The 2019 Stack Overflow Developer Survey Results Are InImproper integral $int^pi/2_0 (operatornamecsc x - frac1x),mathrm dx$Improper integral:$ int_-1^2 frac1xdx$ diverges$int_-infty^infty x^2 cdot frac12Ce^,dx$ where C is a constantConvergence of $int_-1^1fracx-1x^5/3dx$Convergence of $int_0^1 fracsqrt e^2+x^2 - e^cos xtan^axdx$Differing (divergent) limits in the improper integral $int_-1^3x^-3dx$Calculate $int_-infty^+inftyfracx1+x^2dx$, what is wrong with this?$int_-1^0 frac(e^1/x)x^2 dx$ Improper integralsEvaluating $int_-infty^infty 1-e^-frac1x^2rm dx$The convergence of a strange integral: $int_0^1fracx-1ln xmathbb dx$
$begingroup$
I broke the improper integral into:
$$lim_ato0 int_-1^a-0 fracdxx^4 +lim_ato0 int_0+a^1 fracdxx^4,$$
as $0$ is the point of discontinuity. Now, while plugging the limits of the first integral, I am facing problem as it is getting complex. Is my approach correct?
calculus functions maxima-minima
edited Mar 24 at 5:54
Rócherz
3,0263823
asked Mar 24 at 5:16
Aditya SahaAditya Saha
42
$endgroup$
add a comment |
$begingroup$
I broke the improper integral into:
$$lim_ato0 int_-1^a-0 fracdxx^4 +lim_ato0 int_0+a^1 fracdxx^4,$$
as $0$ is the point of discontinuity. Now, while plugging the limits of the first integral, I am facing problem as it is getting complex. Is my approach correct?
calculus functions maxima-minima
edited Mar 24 at 5:54
Rócherz
3,0263823
asked Mar 24 at 5:16
Aditya SahaAditya Saha
42
$endgroup$
$begingroup$
Complex? How so? The integration should be pretty straight forward. Applying the limits it should be clear that things are going off to infinity, and the integral does not converge.
$endgroup$
– Doug M
Mar 24 at 6:00
$begingroup$
Why you need all this - the integral is well defined. Why you call it improper?
$endgroup$
– Moti
Mar 24 at 6:29
$begingroup$
@Moti - This is improper because of the pole at $x = 0$. This is not allowed in the definition of the Riemann integral.
$endgroup$
– Paul Sinclair
Mar 24 at 15:55
$begingroup$
I have solved it. Sometimes easy thing becomes difficult if we see it in a hard way.
$endgroup$
– Aditya Saha
Mar 25 at 13:55
add a comment |
$begingroup$
I broke the improper integral into:
$$lim_ato0 int_-1^a-0 fracdxx^4 +lim_ato0 int_0+a^1 fracdxx^4,$$
as $0$ is the point of discontinuity. Now, while plugging the limits of the first integral, I am facing problem as it is getting complex. Is my approach correct?
calculus functions maxima-minima
edited Mar 24 at 5:54
Rócherz
3,0263823
asked Mar 24 at 5:16
Aditya SahaAditya Saha
42
$endgroup$
I broke the improper integral into:
$$lim_ato0 int_-1^a-0 fracdxx^4 +lim_ato0 int_0+a^1 fracdxx^4,$$
as $0$ is the point of discontinuity. Now, while plugging the limits of the first integral, I am facing problem as it is getting complex. Is my approach correct?
calculus functions maxima-minima
calculus functions maxima-minima
edited Mar 24 at 5:54
Rócherz
3,0263823
asked Mar 24 at 5:16
Aditya SahaAditya Saha
42
edited Mar 24 at 5:54
Rócherz
3,0263823
asked Mar 24 at 5:16
Aditya SahaAditya Saha
42
edited Mar 24 at 5:54
Rócherz
3,0263823
edited Mar 24 at 5:54
Rócherz
3,0263823
edited Mar 24 at 5:54
Rócherz
3,0263823
3,0263823
asked Mar 24 at 5:16
Aditya SahaAditya Saha
42
asked Mar 24 at 5:16
Aditya SahaAditya Saha
42
asked Mar 24 at 5:16
Aditya SahaAditya Saha
42
42
$begingroup$
Complex? How so? The integration should be pretty straight forward. Applying the limits it should be clear that things are going off to infinity, and the integral does not converge.
$endgroup$
– Doug M
Mar 24 at 6:00
$begingroup$
Why you need all this - the integral is well defined. Why you call it improper?
$endgroup$
– Moti
Mar 24 at 6:29
$begingroup$
@Moti - This is improper because of the pole at $x = 0$. This is not allowed in the definition of the Riemann integral.
$endgroup$
– Paul Sinclair
Mar 24 at 15:55
$begingroup$
I have solved it. Sometimes easy thing becomes difficult if we see it in a hard way.
$endgroup$
– Aditya Saha
Mar 25 at 13:55
add a comment |
$begingroup$
Complex? How so? The integration should be pretty straight forward. Applying the limits it should be clear that things are going off to infinity, and the integral does not converge.
$endgroup$
– Doug M
Mar 24 at 6:00
$begingroup$
Why you need all this - the integral is well defined. Why you call it improper?
$endgroup$
– Moti
Mar 24 at 6:29
$begingroup$
@Moti - This is improper because of the pole at $x = 0$. This is not allowed in the definition of the Riemann integral.
$endgroup$
– Paul Sinclair
Mar 24 at 15:55
$begingroup$
I have solved it. Sometimes easy thing becomes difficult if we see it in a hard way.
$endgroup$
– Aditya Saha
Mar 25 at 13:55
$begingroup$
Complex? How so? The integration should be pretty straight forward. Applying the limits it should be clear that things are going off to infinity, and the integral does not converge.
$endgroup$
– Doug M
Mar 24 at 6:00
$begingroup$
Complex? How so? The integration should be pretty straight forward. Applying the limits it should be clear that things are going off to infinity, and the integral does not converge.
$endgroup$
– Doug M
Mar 24 at 6:00
$begingroup$
Why you need all this - the integral is well defined. Why you call it improper?
$endgroup$
– Moti
Mar 24 at 6:29
$begingroup$
Why you need all this - the integral is well defined. Why you call it improper?
$endgroup$
– Moti
Mar 24 at 6:29
$begingroup$
@Moti - This is improper because of the pole at $x = 0$. This is not allowed in the definition of the Riemann integral.
$endgroup$
– Paul Sinclair
Mar 24 at 15:55
$begingroup$
@Moti - This is improper because of the pole at $x = 0$. This is not allowed in the definition of the Riemann integral.
$endgroup$
– Paul Sinclair
Mar 24 at 15:55
$begingroup$
I have solved it. Sometimes easy thing becomes difficult if we see it in a hard way.
$endgroup$
– Aditya Saha
Mar 25 at 13:55
$begingroup$
I have solved it. Sometimes easy thing becomes difficult if we see it in a hard way.
$endgroup$
– Aditya Saha
Mar 25 at 13:55
add a comment |
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active
oldest
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$begingroup$
Complex? How so? The integration should be pretty straight forward. Applying the limits it should be clear that things are going off to infinity, and the integral does not converge.
$endgroup$
– Doug M
Mar 24 at 6:00
$begingroup$
Why you need all this - the integral is well defined. Why you call it improper?
$endgroup$
– Moti
Mar 24 at 6:29
$begingroup$
@Moti - This is improper because of the pole at $x = 0$. This is not allowed in the definition of the Riemann integral.
$endgroup$
– Paul Sinclair
Mar 24 at 15:55
$begingroup$
I have solved it. Sometimes easy thing becomes difficult if we see it in a hard way.
$endgroup$
– Aditya Saha
Mar 25 at 13:55