Integral $int_-1^1 fracmathrm dxx$Evaluating the definite integral $int_0^infty x mathrme^-frac(x-a)^2b,mathrmdx$$int_Ofracpartial^2 wpartial x^2cdotfracpartial^2 wpartial y^2=int_O(fracpartial^2 wpartial xpartial y)^2$Does this integral converge? $int_0^infty e^largex/2-2alpha(x^2-x^1+delta)mathrm dx$Changing integration order in $int_0^4int_0^1int_2y^2 frac4cos (x^2)2sqrtz,mathrmdx,mathrmdy,mathrmdz$Integral whose upper limit is the integral itself: $int_0^int_0^ldotsfrac1sqrtx mathrmdx frac1sqrtx mathrmdx$Evaluate $int_1^+inftye^-x,fracsin(alpha x)x,mathrmdx$Convergence of improper Integral $int_0^infty cos (x^3)mathrmdx$Give a recursive formula for calculating the value of $ int_0^pi sin^n x , mathrm dx$Evaluating a definite integral: $int_0^pifracx1-sinxcosx,mathrm dx$Gaussian type integral $int_-infty^infty fracmathrme^-a^2 x^21 + x^2 mathrmdx$
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Integral $int_-1^1 fracmathrm dxx$
Evaluating the definite integral $int_0^infty x mathrme^-frac(x-a)^2b,mathrmdx$$int_Ofracpartial^2 wpartial x^2cdotfracpartial^2 wpartial y^2=int_O(fracpartial^2 wpartial xpartial y)^2$Does this integral converge? $int_0^infty e^largex/2-2alpha(x^2-x^1+delta)mathrm dx$Changing integration order in $int_0^4int_0^1int_2y^2 frac4cos (x^2)2sqrtz,mathrmdx,mathrmdy,mathrmdz$Integral whose upper limit is the integral itself: $int_0^int_0^ldotsfrac1sqrtx mathrmdx frac1sqrtx mathrmdx$Evaluate $int_1^+inftye^-x,fracsin(alpha x)x,mathrmdx$Convergence of improper Integral $int_0^infty cos (x^3)mathrmdx$Give a recursive formula for calculating the value of $ int_0^pi sin^n x , mathrm dx$Evaluating a definite integral: $int_0^pifracx1-sinxcosx,mathrm dx$Gaussian type integral $int_-infty^infty fracmathrme^-a^2 x^21 + x^2 mathrmdx$
$begingroup$
Consider the following integration
$$int_-1^1fracmathrm dxx$$
B thinks this expression should be written as
$$int_-1^0fracmathrm dxx+int_0^1fracmathrm dxx$$
which is not convergent, but A thinks it is zero because it is an odd function with the symmetric region so the integration is zero. B thinks that the integration doesn't converge, so we are not allowed to use the properties of integration.
I think B is correct. Right?
calculus
edited May 30 at 8:19
YuiTo Cheng
3,35571545
asked May 29 at 23:09
STUDENTSTUDENT
1527
$endgroup$
add a comment |
$begingroup$
Consider the following integration
$$int_-1^1fracmathrm dxx$$
B thinks this expression should be written as
$$int_-1^0fracmathrm dxx+int_0^1fracmathrm dxx$$
which is not convergent, but A thinks it is zero because it is an odd function with the symmetric region so the integration is zero. B thinks that the integration doesn't converge, so we are not allowed to use the properties of integration.
I think B is correct. Right?
calculus
edited May 30 at 8:19
YuiTo Cheng
3,35571545
asked May 29 at 23:09
STUDENTSTUDENT
1527
$endgroup$
$begingroup$
don't forget $dx$
$endgroup$
– J. W. Tanner
May 29 at 23:15
add a comment |
$begingroup$
Consider the following integration
$$int_-1^1fracmathrm dxx$$
B thinks this expression should be written as
$$int_-1^0fracmathrm dxx+int_0^1fracmathrm dxx$$
which is not convergent, but A thinks it is zero because it is an odd function with the symmetric region so the integration is zero. B thinks that the integration doesn't converge, so we are not allowed to use the properties of integration.
I think B is correct. Right?
calculus
edited May 30 at 8:19
YuiTo Cheng
3,35571545
asked May 29 at 23:09
STUDENTSTUDENT
1527
$endgroup$
Consider the following integration
$$int_-1^1fracmathrm dxx$$
B thinks this expression should be written as
$$int_-1^0fracmathrm dxx+int_0^1fracmathrm dxx$$
which is not convergent, but A thinks it is zero because it is an odd function with the symmetric region so the integration is zero. B thinks that the integration doesn't converge, so we are not allowed to use the properties of integration.
I think B is correct. Right?
calculus
calculus
edited May 30 at 8:19
YuiTo Cheng
3,35571545
asked May 29 at 23:09
STUDENTSTUDENT
1527
edited May 30 at 8:19
YuiTo Cheng
3,35571545
asked May 29 at 23:09
STUDENTSTUDENT
1527
edited May 30 at 8:19
YuiTo Cheng
3,35571545
edited May 30 at 8:19
YuiTo Cheng
3,35571545
edited May 30 at 8:19
YuiTo Cheng
3,35571545
3,35571545
asked May 29 at 23:09
STUDENTSTUDENT
1527
asked May 29 at 23:09
STUDENTSTUDENT
1527
asked May 29 at 23:09
STUDENTSTUDENT
1527
1527
$begingroup$
don't forget $dx$
$endgroup$
– J. W. Tanner
May 29 at 23:15
add a comment |
$begingroup$
don't forget $dx$
$endgroup$
– J. W. Tanner
May 29 at 23:15
$begingroup$
don't forget $dx$
$endgroup$
– J. W. Tanner
May 29 at 23:15
$begingroup$
don't forget $dx$
$endgroup$
– J. W. Tanner
May 29 at 23:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Student B is correct in the sense that traditionally, this integral cannot be evaluated. However, the Cauchy principal value of this integral exists (the 'intuitive' way to extend integrals into areas where they do not converge) and is indeed 0.
answered May 29 at 23:12
auscryptauscrypt
5,186212
$endgroup$
add a comment |
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$begingroup$
Student B is correct in the sense that traditionally, this integral cannot be evaluated. However, the Cauchy principal value of this integral exists (the 'intuitive' way to extend integrals into areas where they do not converge) and is indeed 0.
answered May 29 at 23:12
auscryptauscrypt
5,186212
$endgroup$
add a comment |
$begingroup$
Student B is correct in the sense that traditionally, this integral cannot be evaluated. However, the Cauchy principal value of this integral exists (the 'intuitive' way to extend integrals into areas where they do not converge) and is indeed 0.
answered May 29 at 23:12
auscryptauscrypt
5,186212
$endgroup$
add a comment |
$begingroup$
Student B is correct in the sense that traditionally, this integral cannot be evaluated. However, the Cauchy principal value of this integral exists (the 'intuitive' way to extend integrals into areas where they do not converge) and is indeed 0.
answered May 29 at 23:12
auscryptauscrypt
5,186212
$endgroup$
Student B is correct in the sense that traditionally, this integral cannot be evaluated. However, the Cauchy principal value of this integral exists (the 'intuitive' way to extend integrals into areas where they do not converge) and is indeed 0.
answered May 29 at 23:12
auscryptauscrypt
5,186212
answered May 29 at 23:12
auscryptauscrypt
5,186212
answered May 29 at 23:12
auscryptauscrypt
5,186212
answered May 29 at 23:12
auscryptauscrypt
5,186212
5,186212
add a comment |
add a comment |
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$begingroup$
don't forget $dx$
$endgroup$
– J. W. Tanner
May 29 at 23:15