I am getting undefined as the answer of this integral problem $intlimits_2^3fracmathrm dn(n-2)(3-n)$. Am I doing something wrong? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Improper integral $int^pi/2_0 (operatornamecsc x - frac1x),mathrm dx$an intriguing integral $I=intlimits_0^4 fracdx4+2^x $Computing the integral $intlimits_-infty^infty(t^2-1)delta(t):dt$Help to understand this property: $intlimits_ka^kbsleft(fracxkright)dx = kintlimits_a^bs(x)dx$Prove $intlimits_0^infty mathrmexp(-ax^2-fracbx^2) mathrmd x = frac12sqrtfracpia}mathrme^{-2sqrtab$Computing the integral $int fracdxxsqrt x^2+1$Why am I not getting the right answer for this integral?Show that $intlimits_0^fracpi24cos^2(x)log^2(cos x)~mathrm dx=-pilog 2+pilog^2 2-fracpi2+fracpi^312$Calculating improper integral $int limits_0^inftyfracmathrme^-xsqrtx,mathrmdx$Finding the value of $limlimits_nrightarrow inftysqrtnint^fracpi4_0cos^2n-2(x)mathrm dx$
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I am getting undefined as the answer of this integral problem $intlimits_2^3fracmathrm dn(n-2)(3-n)$. Am I doing something wrong?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Improper integral $int^pi/2_0 (operatornamecsc x - frac1x),mathrm dx$an intriguing integral $I=intlimits_0^4 fracdx4+2^x $Computing the integral $intlimits_-infty^infty(t^2-1)delta(t):dt$Help to understand this property: $intlimits_ka^kbsleft(fracxkright)dx = kintlimits_a^bs(x)dx$Prove $intlimits_0^infty mathrmexp(-ax^2-fracbx^2) mathrmd x = frac12sqrtfracpiamathrme^-2sqrtab$Computing the integral $int fracdxxsqrt x^2+1$Why am I not getting the right answer for this integral?Show that $intlimits_0^fracpi24cos^2(x)log^2(cos x)~mathrm dx=-pilog 2+pilog^2 2-fracpi2+fracpi^312$Calculating improper integral $int limits_0^inftyfracmathrme^-xsqrtx,mathrmdx$Finding the value of $limlimits_nrightarrow inftysqrtnint^fracpi4_0cos^2n-2(x)mathrm dx$
$begingroup$
Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$
My Attempt:
Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(3-n)+B(n-2)(n-2)(3-n)\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-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]endalign$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
edited Apr 14 at 11:59
YuiTo Cheng
2,54841037
asked Apr 14 at 6:46
arandomguyarandomguy
17418
$endgroup$
add a comment |
$begingroup$
Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$
My Attempt:
Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(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.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-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]endalign$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
edited Apr 14 at 11:59
YuiTo Cheng
2,54841037
asked Apr 14 at 6:46
arandomguyarandomguy
17418
$endgroup$
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
Apr 14 at 7:01
add a comment |
$begingroup$
Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$
My Attempt:
Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(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.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-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]endalign$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
integration definite-integrals
edited Apr 14 at 11:59
YuiTo Cheng
2,54841037
asked Apr 14 at 6:46
arandomguyarandomguy
17418
$endgroup$
Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$
My Attempt:
Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(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.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-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]endalign$$ 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 Apr 14 at 11:59
YuiTo Cheng
2,54841037
asked Apr 14 at 6:46
arandomguyarandomguy
17418
edited Apr 14 at 11:59
YuiTo Cheng
2,54841037
asked Apr 14 at 6:46
arandomguyarandomguy
17418
edited Apr 14 at 11:59
YuiTo Cheng
2,54841037
edited Apr 14 at 11:59
YuiTo Cheng
2,54841037
edited Apr 14 at 11:59
YuiTo Cheng
2,54841037
2,54841037
asked Apr 14 at 6:46
arandomguyarandomguy
17418
asked Apr 14 at 6:46
arandomguyarandomguy
17418
asked Apr 14 at 6:46
arandomguyarandomguy
17418
17418
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
Apr 14 at 7:01
add a comment |
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
Apr 14 at 7:01
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
Apr 14 at 7:01
$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
Apr 14 at 7:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Apr 14 at 7:09
The_SympathizerThe_Sympathizer
7,8952246
$endgroup$
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
Apr 14 at 7:09
$begingroup$
@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
Apr 14 at 7:12
$begingroup$
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
Apr 14 at 7:14
add a comment |
$begingroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered Apr 14 at 7:09
José Carlos SantosJosé Carlos Santos
176k24134243
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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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.
answered Apr 14 at 7:09
The_SympathizerThe_Sympathizer
7,8952246
$endgroup$
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
Apr 14 at 7:09
$begingroup$
@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
Apr 14 at 7:12
$begingroup$
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
Apr 14 at 7:14
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.
answered Apr 14 at 7:09
The_SympathizerThe_Sympathizer
7,8952246
$endgroup$
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
Apr 14 at 7:09
$begingroup$
@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
Apr 14 at 7:12
$begingroup$
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
Apr 14 at 7:14
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.
answered Apr 14 at 7:09
The_SympathizerThe_Sympathizer
7,8952246
$endgroup$
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 Apr 14 at 7:09
The_SympathizerThe_Sympathizer
7,8952246
answered Apr 14 at 7:09
The_SympathizerThe_Sympathizer
7,8952246
answered Apr 14 at 7:09
The_SympathizerThe_Sympathizer
7,8952246
answered Apr 14 at 7:09
The_SympathizerThe_Sympathizer
7,8952246
7,8952246
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
Apr 14 at 7:09
$begingroup$
@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
Apr 14 at 7:12
$begingroup$
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
Apr 14 at 7:14
add a comment |
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
Apr 14 at 7:09
$begingroup$
@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
Apr 14 at 7:12
$begingroup$
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
Apr 14 at 7:14
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
Apr 14 at 7:09
$begingroup$
Whats quite strong mean
$endgroup$
– Mikey Spivak
Apr 14 at 7:09
$begingroup$
@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
Apr 14 at 7:12
$begingroup$
@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
Apr 14 at 7:12
$begingroup$
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
Apr 14 at 7:14
$begingroup$
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
Apr 14 at 7:14
add a comment |
$begingroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered Apr 14 at 7:09
José Carlos SantosJosé Carlos Santos
176k24134243
$endgroup$
add a comment |
$begingroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered Apr 14 at 7:09
José Carlos SantosJosé Carlos Santos
176k24134243
$endgroup$
add a comment |
$begingroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered Apr 14 at 7:09
José Carlos SantosJosé Carlos Santos
176k24134243
$endgroup$
No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.
answered Apr 14 at 7:09
José Carlos SantosJosé Carlos Santos
176k24134243
answered Apr 14 at 7:09
José Carlos SantosJosé Carlos Santos
176k24134243
answered Apr 14 at 7:09
José Carlos SantosJosé Carlos Santos
176k24134243
answered Apr 14 at 7:09
José Carlos SantosJosé Carlos Santos
176k24134243
176k24134243
add a comment |
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
Apr 14 at 7:01