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Multiple answers in evaluation of a definite integral.
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Consider this definite integral
$$I=int_0^2pidfracxcos x1 +cos xdx$$
Method 1
$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$
$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$
Adding (1) and (2)
$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$
$$I=pi int_0^2pidfraccos x1 +cos xdx$$
$$I=2pi int_0^pidfraccos x1 +cos xdx.....(P1)$$
$$I=2piint_0^pidfrac1+cos x-11 +cos xdx$$
$$I=2pi int_0^pi1-dfrac11 +cos xdx$$
$$I=2pi int_0^pi1dx-2pi int_0^pidfrac11 +cos xdx$$
$$I=2pi[x]_0^pi -2pi int_0^pidfrac12cos ²dfracx2dx$$
$$I=2pi ² - pi int_0^pi sec ²dfracx2 dx$$
$$displaystyle I=2pi ² - 2pi left[ tan dfracx2right]_0^pi $$
$$displaystyle I=2pi ² - 2pi left[ infty - 0 right] $$
$$I=-infty$$
Method 2
$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$
$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$
Adding (1) and (2)
$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$
$$I=pi int_0^2pidfraccos x1 +cos xdx$$
$$I=piint_0^2pidfrac1+cos x-11 +cos xdx$$
$$I=pi int_0^2pi1-dfrac11 +cos xdx$$
$$I=pi int_0^2pi1dx-pi int_0^2pidfrac11 +cos xdx$$
$$I=pi[x]_0^2pi -pi int_0^2pidfrac12cos ²dfracx2dx$$
$$I=2pi ² - dfracpi2 int_0^2pi sec ²dfracx2 dx$$
$$displaystyle I=2pi ² - pi left[ tan dfracx2right]_0^2pi $$
$$displaystyle I=2pi ² - pi left[ 0-0right] $$
$$I=2pi ²$$
$$(P1)--> int_0^2a f(x)dx=2int_0^af(x)dx$$if $f(2a-x)=f(x)$
I have checked these methods several times, but can't find any mistake. Method 1 seems to be wrong as the given answer was $2pi ²$. What could be the mistake?
definite-integrals
asked 1 hour ago
Loop Back
2518
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up vote
1
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Consider this definite integral
$$I=int_0^2pidfracxcos x1 +cos xdx$$
Method 1
$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$
$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$
Adding (1) and (2)
$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$
$$I=pi int_0^2pidfraccos x1 +cos xdx$$
$$I=2pi int_0^pidfraccos x1 +cos xdx.....(P1)$$
$$I=2piint_0^pidfrac1+cos x-11 +cos xdx$$
$$I=2pi int_0^pi1-dfrac11 +cos xdx$$
$$I=2pi int_0^pi1dx-2pi int_0^pidfrac11 +cos xdx$$
$$I=2pi[x]_0^pi -2pi int_0^pidfrac12cos ²dfracx2dx$$
$$I=2pi ² - pi int_0^pi sec ²dfracx2 dx$$
$$displaystyle I=2pi ² - 2pi left[ tan dfracx2right]_0^pi $$
$$displaystyle I=2pi ² - 2pi left[ infty - 0 right] $$
$$I=-infty$$
Method 2
$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$
$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$
Adding (1) and (2)
$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$
$$I=pi int_0^2pidfraccos x1 +cos xdx$$
$$I=piint_0^2pidfrac1+cos x-11 +cos xdx$$
$$I=pi int_0^2pi1-dfrac11 +cos xdx$$
$$I=pi int_0^2pi1dx-pi int_0^2pidfrac11 +cos xdx$$
$$I=pi[x]_0^2pi -pi int_0^2pidfrac12cos ²dfracx2dx$$
$$I=2pi ² - dfracpi2 int_0^2pi sec ²dfracx2 dx$$
$$displaystyle I=2pi ² - pi left[ tan dfracx2right]_0^2pi $$
$$displaystyle I=2pi ² - pi left[ 0-0right] $$
$$I=2pi ²$$
$$(P1)--> int_0^2a f(x)dx=2int_0^af(x)dx$$if $f(2a-x)=f(x)$
I have checked these methods several times, but can't find any mistake. Method 1 seems to be wrong as the given answer was $2pi ²$. What could be the mistake?
definite-integrals
asked 1 hour ago
Loop Back
2518
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider this definite integral
$$I=int_0^2pidfracxcos x1 +cos xdx$$
Method 1
$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$
$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$
Adding (1) and (2)
$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$
$$I=pi int_0^2pidfraccos x1 +cos xdx$$
$$I=2pi int_0^pidfraccos x1 +cos xdx.....(P1)$$
$$I=2piint_0^pidfrac1+cos x-11 +cos xdx$$
$$I=2pi int_0^pi1-dfrac11 +cos xdx$$
$$I=2pi int_0^pi1dx-2pi int_0^pidfrac11 +cos xdx$$
$$I=2pi[x]_0^pi -2pi int_0^pidfrac12cos ²dfracx2dx$$
$$I=2pi ² - pi int_0^pi sec ²dfracx2 dx$$
$$displaystyle I=2pi ² - 2pi left[ tan dfracx2right]_0^pi $$
$$displaystyle I=2pi ² - 2pi left[ infty - 0 right] $$
$$I=-infty$$
Method 2
$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$
$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$
Adding (1) and (2)
$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$
$$I=pi int_0^2pidfraccos x1 +cos xdx$$
$$I=piint_0^2pidfrac1+cos x-11 +cos xdx$$
$$I=pi int_0^2pi1-dfrac11 +cos xdx$$
$$I=pi int_0^2pi1dx-pi int_0^2pidfrac11 +cos xdx$$
$$I=pi[x]_0^2pi -pi int_0^2pidfrac12cos ²dfracx2dx$$
$$I=2pi ² - dfracpi2 int_0^2pi sec ²dfracx2 dx$$
$$displaystyle I=2pi ² - pi left[ tan dfracx2right]_0^2pi $$
$$displaystyle I=2pi ² - pi left[ 0-0right] $$
$$I=2pi ²$$
$$(P1)--> int_0^2a f(x)dx=2int_0^af(x)dx$$if $f(2a-x)=f(x)$
I have checked these methods several times, but can't find any mistake. Method 1 seems to be wrong as the given answer was $2pi ²$. What could be the mistake?
definite-integrals
asked 1 hour ago
Loop Back
2518
Consider this definite integral
$$I=int_0^2pidfracxcos x1 +cos xdx$$
Method 1
$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$
$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$
Adding (1) and (2)
$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$
$$I=pi int_0^2pidfraccos x1 +cos xdx$$
$$I=2pi int_0^pidfraccos x1 +cos xdx.....(P1)$$
$$I=2piint_0^pidfrac1+cos x-11 +cos xdx$$
$$I=2pi int_0^pi1-dfrac11 +cos xdx$$
$$I=2pi int_0^pi1dx-2pi int_0^pidfrac11 +cos xdx$$
$$I=2pi[x]_0^pi -2pi int_0^pidfrac12cos ²dfracx2dx$$
$$I=2pi ² - pi int_0^pi sec ²dfracx2 dx$$
$$displaystyle I=2pi ² - 2pi left[ tan dfracx2right]_0^pi $$
$$displaystyle I=2pi ² - 2pi left[ infty - 0 right] $$
$$I=-infty$$
Method 2
$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$
$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$
Adding (1) and (2)
$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$
$$I=pi int_0^2pidfraccos x1 +cos xdx$$
$$I=piint_0^2pidfrac1+cos x-11 +cos xdx$$
$$I=pi int_0^2pi1-dfrac11 +cos xdx$$
$$I=pi int_0^2pi1dx-pi int_0^2pidfrac11 +cos xdx$$
$$I=pi[x]_0^2pi -pi int_0^2pidfrac12cos ²dfracx2dx$$
$$I=2pi ² - dfracpi2 int_0^2pi sec ²dfracx2 dx$$
$$displaystyle I=2pi ² - pi left[ tan dfracx2right]_0^2pi $$
$$displaystyle I=2pi ² - pi left[ 0-0right] $$
$$I=2pi ²$$
$$(P1)--> int_0^2a f(x)dx=2int_0^af(x)dx$$if $f(2a-x)=f(x)$
I have checked these methods several times, but can't find any mistake. Method 1 seems to be wrong as the given answer was $2pi ²$. What could be the mistake?
definite-integrals
definite-integrals
asked 1 hour ago
Loop Back
2518
asked 1 hour ago
Loop Back
2518
asked 1 hour ago
Loop Back
2518
asked 1 hour ago
Loop Back
2518
asked 1 hour ago
Loop Back
2518
2518
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
Actually, the first method is correct and the second method is wrong. To see what is happening here, notice that
$$ lim_x to pi (x-pi)^2 fracx cos x1+cos x
= -2pi, $$
meaning that the integrand has pole of order $2$ at $pi$.
$hspace8.5em$
So the integral does not converge (not even in ordinary sense, but as in improper sense and as in Cauchy principal value sense).
To see what is wrong in your second approach, notice that $u = tan(x/2)$ has pole at $pi$, hence cannot be used as a valid substitution over the interval $[0, 2pi]$. (Of course, it can be used as a substitution over the interval $[0,pi)$, which is why your first approach is valid.)
$hspace8.5em$
answered 54 mins ago
Sangchul Lee
86.7k12155255
$SangchulLee thank you I didn't notice that substitution mistake, next time I'll remember it.
– Loop Back
40 mins ago
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
Actually, the first method is correct and the second method is wrong. To see what is happening here, notice that
$$ lim_x to pi (x-pi)^2 fracx cos x1+cos x
= -2pi, $$
meaning that the integrand has pole of order $2$ at $pi$.
$hspace8.5em$
So the integral does not converge (not even in ordinary sense, but as in improper sense and as in Cauchy principal value sense).
To see what is wrong in your second approach, notice that $u = tan(x/2)$ has pole at $pi$, hence cannot be used as a valid substitution over the interval $[0, 2pi]$. (Of course, it can be used as a substitution over the interval $[0,pi)$, which is why your first approach is valid.)
$hspace8.5em$
answered 54 mins ago
Sangchul Lee
86.7k12155255
$SangchulLee thank you I didn't notice that substitution mistake, next time I'll remember it.
– Loop Back
40 mins ago
add a comment |Â
up vote
6
down vote
accepted
Actually, the first method is correct and the second method is wrong. To see what is happening here, notice that
$$ lim_x to pi (x-pi)^2 fracx cos x1+cos x
= -2pi, $$
meaning that the integrand has pole of order $2$ at $pi$.
$hspace8.5em$
So the integral does not converge (not even in ordinary sense, but as in improper sense and as in Cauchy principal value sense).
To see what is wrong in your second approach, notice that $u = tan(x/2)$ has pole at $pi$, hence cannot be used as a valid substitution over the interval $[0, 2pi]$. (Of course, it can be used as a substitution over the interval $[0,pi)$, which is why your first approach is valid.)
$hspace8.5em$
answered 54 mins ago
Sangchul Lee
86.7k12155255
$SangchulLee thank you I didn't notice that substitution mistake, next time I'll remember it.
– Loop Back
40 mins ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Actually, the first method is correct and the second method is wrong. To see what is happening here, notice that
$$ lim_x to pi (x-pi)^2 fracx cos x1+cos x
= -2pi, $$
meaning that the integrand has pole of order $2$ at $pi$.
$hspace8.5em$
So the integral does not converge (not even in ordinary sense, but as in improper sense and as in Cauchy principal value sense).
To see what is wrong in your second approach, notice that $u = tan(x/2)$ has pole at $pi$, hence cannot be used as a valid substitution over the interval $[0, 2pi]$. (Of course, it can be used as a substitution over the interval $[0,pi)$, which is why your first approach is valid.)
$hspace8.5em$
answered 54 mins ago
Sangchul Lee
86.7k12155255
Actually, the first method is correct and the second method is wrong. To see what is happening here, notice that
$$ lim_x to pi (x-pi)^2 fracx cos x1+cos x
= -2pi, $$
meaning that the integrand has pole of order $2$ at $pi$.
$hspace8.5em$
So the integral does not converge (not even in ordinary sense, but as in improper sense and as in Cauchy principal value sense).
To see what is wrong in your second approach, notice that $u = tan(x/2)$ has pole at $pi$, hence cannot be used as a valid substitution over the interval $[0, 2pi]$. (Of course, it can be used as a substitution over the interval $[0,pi)$, which is why your first approach is valid.)
$hspace8.5em$
answered 54 mins ago
Sangchul Lee
86.7k12155255
answered 54 mins ago
Sangchul Lee
86.7k12155255
answered 54 mins ago
Sangchul Lee
86.7k12155255
answered 54 mins ago
Sangchul Lee
86.7k12155255
86.7k12155255
$SangchulLee thank you I didn't notice that substitution mistake, next time I'll remember it.
– Loop Back
40 mins ago
add a comment |Â
$SangchulLee thank you I didn't notice that substitution mistake, next time I'll remember it.
– Loop Back
40 mins ago
$SangchulLee thank you I didn't notice that substitution mistake, next time I'll remember it.
– Loop Back
40 mins ago
$SangchulLee thank you I didn't notice that substitution mistake, next time I'll remember it.
– Loop Back
40 mins ago
add a comment |Â
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