I am getting undefined as the answer of this integral problem $intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$....
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Find $$intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$$
My Attempt:
Let $$begin{align}frac{1}{(n-2)(3-n)}&=frac{A}{n-2}+frac{B}{3-n} \ &= frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$begin{align}therefore intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)} &= intlimits_{2}^{3}bigg(frac{1}{n-2}+frac{1}{3-n}bigg),mathrm dn \ &= intlimits_{2}^{3}frac{1}{n-2},mathrm dn+intlimits_{2}^{3}frac{1}{3-n},mathrm dn \ &=left[log(n-2)right]_{small 2}^{small 3}+left[log(3-n)right]_{small 2}^{small 3} \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]end{align}$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
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add a comment |
$begingroup$
Find $$intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$$
My Attempt:
Let $$begin{align}frac{1}{(n-2)(3-n)}&=frac{A}{n-2}+frac{B}{3-n} \ &= frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$begin{align}therefore intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)} &= intlimits_{2}^{3}bigg(frac{1}{n-2}+frac{1}{3-n}bigg),mathrm dn \ &= intlimits_{2}^{3}frac{1}{n-2},mathrm dn+intlimits_{2}^{3}frac{1}{3-n},mathrm dn \ &=left[log(n-2)right]_{small 2}^{small 3}+left[log(3-n)right]_{small 2}^{small 3} \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]end{align}$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
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Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
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– Dbchatto67
20 hours ago
add a comment |
$begingroup$
Find $$intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$$
My Attempt:
Let $$begin{align}frac{1}{(n-2)(3-n)}&=frac{A}{n-2}+frac{B}{3-n} \ &= frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$begin{align}therefore intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)} &= intlimits_{2}^{3}bigg(frac{1}{n-2}+frac{1}{3-n}bigg),mathrm dn \ &= intlimits_{2}^{3}frac{1}{n-2},mathrm dn+intlimits_{2}^{3}frac{1}{3-n},mathrm dn \ &=left[log(n-2)right]_{small 2}^{small 3}+left[log(3-n)right]_{small 2}^{small 3} \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]end{align}$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
$endgroup$
Find $$intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$$
My Attempt:
Let $$begin{align}frac{1}{(n-2)(3-n)}&=frac{A}{n-2}+frac{B}{3-n} \ &= frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$begin{align}therefore intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)} &= intlimits_{2}^{3}bigg(frac{1}{n-2}+frac{1}{3-n}bigg),mathrm dn \ &= intlimits_{2}^{3}frac{1}{n-2},mathrm dn+intlimits_{2}^{3}frac{1}{3-n},mathrm dn \ &=left[log(n-2)right]_{small 2}^{small 3}+left[log(3-n)right]_{small 2}^{small 3} \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]end{align}$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
integration definite-integrals
edited 15 hours ago
YuiTo Cheng
2,41341037
2,41341037
asked 20 hours ago
arandomguyarandomguy
17418
17418
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Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
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– Dbchatto67
20 hours ago
add a comment |
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
20 hours ago
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
20 hours ago
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
20 hours ago
add a comment |
2 Answers
2
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No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
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Whats quite strong mean
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– Mikey Spivak
20 hours ago
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@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
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– The_Sympathizer
20 hours ago
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More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
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– The_Sympathizer
20 hours ago
add a comment |
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No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.
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Your Answer
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2 Answers
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active
oldest
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2 Answers
2
active
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active
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active
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$begingroup$
No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
$endgroup$
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
20 hours ago
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
20 hours ago
$begingroup$
More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
20 hours ago
add a comment |
$begingroup$
No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
$endgroup$
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
20 hours ago
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
20 hours ago
$begingroup$
More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
20 hours ago
add a comment |
$begingroup$
No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
$endgroup$
No.
There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.
answered 20 hours ago
The_SympathizerThe_Sympathizer
7,8752246
7,8752246
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
20 hours ago
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
20 hours ago
$begingroup$
More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
20 hours ago
add a comment |
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
20 hours ago
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
20 hours ago
$begingroup$
More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
20 hours ago
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
20 hours ago
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
20 hours ago
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
20 hours ago
$begingroup$
@MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
$endgroup$
– The_Sympathizer
20 hours ago
$begingroup$
More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
20 hours ago
$begingroup$
More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
$endgroup$
– The_Sympathizer
20 hours ago
add a comment |
$begingroup$
No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.
$endgroup$
add a comment |
$begingroup$
No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.
$endgroup$
add a comment |
$begingroup$
No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.
$endgroup$
No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.
answered 20 hours ago
José Carlos SantosJosé Carlos Santos
175k23134243
175k23134243
add a comment |
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$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
20 hours ago