Evaluate the integral $int frac {1}{x^4 +1} dx$( finding the constants in the partial fraction method).
up vote
0
down vote
favorite
Evaluate the integral $int frac {1}{x^4 +1} dx$.
This question is answered here :
Evaluate the following indefinite integral. $intfrac{1}{x^4+1}, dx$
But I do not know how they found the constants in the partial fraction method, could anyone explain this for me please?
calculus real-analysis integration analysis
asked Nov 19 at 9:35
hopefully
132112
add a comment |
up vote
0
down vote
favorite
Evaluate the integral $int frac {1}{x^4 +1} dx$.
This question is answered here :
Evaluate the following indefinite integral. $intfrac{1}{x^4+1}, dx$
But I do not know how they found the constants in the partial fraction method, could anyone explain this for me please?
calculus real-analysis integration analysis
asked Nov 19 at 9:35
hopefully
132112
2
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
Nov 19 at 10:16
Evaluation of integral of $frac{1}{x^4 + 1}$ AND evaluation of integral of $frac{1}{x^5 + 1}$ AND evaluation of integral of $frac{1}{x^6 + 1}$. See this also.
– Dave L. Renfro
Nov 19 at 13:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Evaluate the integral $int frac {1}{x^4 +1} dx$.
This question is answered here :
Evaluate the following indefinite integral. $intfrac{1}{x^4+1}, dx$
But I do not know how they found the constants in the partial fraction method, could anyone explain this for me please?
calculus real-analysis integration analysis
asked Nov 19 at 9:35
hopefully
132112
Evaluate the integral $int frac {1}{x^4 +1} dx$.
This question is answered here :
Evaluate the following indefinite integral. $intfrac{1}{x^4+1}, dx$
But I do not know how they found the constants in the partial fraction method, could anyone explain this for me please?
calculus real-analysis integration analysis
calculus real-analysis integration analysis
asked Nov 19 at 9:35
hopefully
132112
asked Nov 19 at 9:35
hopefully
132112
asked Nov 19 at 9:35
hopefully
132112
asked Nov 19 at 9:35
hopefully
132112
asked Nov 19 at 9:35
hopefully
132112
132112
2
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
Nov 19 at 10:16
Evaluation of integral of $frac{1}{x^4 + 1}$ AND evaluation of integral of $frac{1}{x^5 + 1}$ AND evaluation of integral of $frac{1}{x^6 + 1}$. See this also.
– Dave L. Renfro
Nov 19 at 13:55
add a comment |
2
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
Nov 19 at 10:16
Evaluation of integral of $frac{1}{x^4 + 1}$ AND evaluation of integral of $frac{1}{x^5 + 1}$ AND evaluation of integral of $frac{1}{x^6 + 1}$. See this also.
– Dave L. Renfro
Nov 19 at 13:55
2
2
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
Nov 19 at 10:16
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
Nov 19 at 10:16
Evaluation of integral of $frac{1}{x^4 + 1}$ AND evaluation of integral of $frac{1}{x^5 + 1}$ AND evaluation of integral of $frac{1}{x^6 + 1}$. See this also.
– Dave L. Renfro
Nov 19 at 13:55
Evaluation of integral of $frac{1}{x^4 + 1}$ AND evaluation of integral of $frac{1}{x^5 + 1}$ AND evaluation of integral of $frac{1}{x^6 + 1}$. See this also.
– Dave L. Renfro
Nov 19 at 13:55
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
If you are doing the integral from $0$ to $infty$ it's worth noting the following. Consider the integral $$frac{1}{2pi i} int frac{log z}{1+z^n} dz, qquad n=4$$ around the classic keyhole contour, that is to say, start at 0 just above the real axis, go along the real axis to $infty$, then anticlockwise in a circle until you get to $infty$ just below the real axis, then back to zero. The log funciton is different on the two sides of the branch cut. The rest is residue calculus. Note that this works for any $nge 2$, and emerges as $(pi/n) mathrm{cosec} (pi/n)$.
A related trick is $$frac{1}{2pi i} int frac{z^a}{1+z^n} dz$$ around the same contour, and let $ato0$ whereupon the branch cut disappears(!) but the answer remains intact.
answered Nov 19 at 10:21
Richard Martin
1,63918
Does this answer the question?
– Szeto
Nov 19 at 12:26
1
Well it does if you are tasked with doing the integral from $0$ to $infty$ and mistakenly assume that the first thing to do must be to figure out the indefinite integral, and then stick the limits in.
– Richard Martin
Nov 19 at 12:30
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
Nov 19 at 13:16
add a comment |
up vote
-1
down vote
You can also do this:
When you have factorised the term $x^4+1$ into factors $p(x) & q(x)$ then
$$ frac{1}{x^4+1} = frac {Ax+M}{p(x)}+frac{Bx+N}{q(x)}$$ and then to find A, B, M and N just take values of x for both sides and equate to find them just like linear equations in two variables.
answered Nov 19 at 9:45
jayant98
35414
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
Nov 19 at 9:46
One of easiest value putting is x=0 and if you can see x=√2 also helps! Also if you see the IIT-JEE papers you will find these types of problem to be solved in less than 5 minutes and the method taught to the aspirants is the same method. It is just the issue of finding so called special x value so that equation is simpler. And for that my friend practice is the best method.
– jayant98
Nov 19 at 9:52
I have tried those values and they do not solve my problem also
– hopefully
Nov 19 at 9:56
$x=0$ implies $1=M+N$, $x = sqrt(2)$ implies $1 = Ax+M + 5Bx+5N$, using $-sqrt(2)$ analogue $1 = -5Ax+5M -Bx+N$, i.e. $Ax+5Bx = -5Ax-Bx$, i.e. $A+B=0$. Now get a handy 4th equation ...
– Stockfish
Nov 19 at 10:49
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If you are doing the integral from $0$ to $infty$ it's worth noting the following. Consider the integral $$frac{1}{2pi i} int frac{log z}{1+z^n} dz, qquad n=4$$ around the classic keyhole contour, that is to say, start at 0 just above the real axis, go along the real axis to $infty$, then anticlockwise in a circle until you get to $infty$ just below the real axis, then back to zero. The log funciton is different on the two sides of the branch cut. The rest is residue calculus. Note that this works for any $nge 2$, and emerges as $(pi/n) mathrm{cosec} (pi/n)$.
A related trick is $$frac{1}{2pi i} int frac{z^a}{1+z^n} dz$$ around the same contour, and let $ato0$ whereupon the branch cut disappears(!) but the answer remains intact.
answered Nov 19 at 10:21
Richard Martin
1,63918
Does this answer the question?
– Szeto
Nov 19 at 12:26
1
Well it does if you are tasked with doing the integral from $0$ to $infty$ and mistakenly assume that the first thing to do must be to figure out the indefinite integral, and then stick the limits in.
– Richard Martin
Nov 19 at 12:30
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
Nov 19 at 13:16
add a comment |
up vote
2
down vote
If you are doing the integral from $0$ to $infty$ it's worth noting the following. Consider the integral $$frac{1}{2pi i} int frac{log z}{1+z^n} dz, qquad n=4$$ around the classic keyhole contour, that is to say, start at 0 just above the real axis, go along the real axis to $infty$, then anticlockwise in a circle until you get to $infty$ just below the real axis, then back to zero. The log funciton is different on the two sides of the branch cut. The rest is residue calculus. Note that this works for any $nge 2$, and emerges as $(pi/n) mathrm{cosec} (pi/n)$.
A related trick is $$frac{1}{2pi i} int frac{z^a}{1+z^n} dz$$ around the same contour, and let $ato0$ whereupon the branch cut disappears(!) but the answer remains intact.
answered Nov 19 at 10:21
Richard Martin
1,63918
Does this answer the question?
– Szeto
Nov 19 at 12:26
1
Well it does if you are tasked with doing the integral from $0$ to $infty$ and mistakenly assume that the first thing to do must be to figure out the indefinite integral, and then stick the limits in.
– Richard Martin
Nov 19 at 12:30
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
Nov 19 at 13:16
add a comment |
up vote
2
down vote
up vote
2
down vote
If you are doing the integral from $0$ to $infty$ it's worth noting the following. Consider the integral $$frac{1}{2pi i} int frac{log z}{1+z^n} dz, qquad n=4$$ around the classic keyhole contour, that is to say, start at 0 just above the real axis, go along the real axis to $infty$, then anticlockwise in a circle until you get to $infty$ just below the real axis, then back to zero. The log funciton is different on the two sides of the branch cut. The rest is residue calculus. Note that this works for any $nge 2$, and emerges as $(pi/n) mathrm{cosec} (pi/n)$.
A related trick is $$frac{1}{2pi i} int frac{z^a}{1+z^n} dz$$ around the same contour, and let $ato0$ whereupon the branch cut disappears(!) but the answer remains intact.
answered Nov 19 at 10:21
Richard Martin
1,63918
If you are doing the integral from $0$ to $infty$ it's worth noting the following. Consider the integral $$frac{1}{2pi i} int frac{log z}{1+z^n} dz, qquad n=4$$ around the classic keyhole contour, that is to say, start at 0 just above the real axis, go along the real axis to $infty$, then anticlockwise in a circle until you get to $infty$ just below the real axis, then back to zero. The log funciton is different on the two sides of the branch cut. The rest is residue calculus. Note that this works for any $nge 2$, and emerges as $(pi/n) mathrm{cosec} (pi/n)$.
A related trick is $$frac{1}{2pi i} int frac{z^a}{1+z^n} dz$$ around the same contour, and let $ato0$ whereupon the branch cut disappears(!) but the answer remains intact.
answered Nov 19 at 10:21
Richard Martin
1,63918
edited Nov 19 at 10:31
answered Nov 19 at 10:21
Richard Martin
1,63918
answered Nov 19 at 10:21
Richard Martin
1,63918
answered Nov 19 at 10:21
Richard Martin
1,63918
1,63918
Does this answer the question?
– Szeto
Nov 19 at 12:26
1
Well it does if you are tasked with doing the integral from $0$ to $infty$ and mistakenly assume that the first thing to do must be to figure out the indefinite integral, and then stick the limits in.
– Richard Martin
Nov 19 at 12:30
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
Nov 19 at 13:16
add a comment |
Does this answer the question?
– Szeto
Nov 19 at 12:26
1
Well it does if you are tasked with doing the integral from $0$ to $infty$ and mistakenly assume that the first thing to do must be to figure out the indefinite integral, and then stick the limits in.
– Richard Martin
Nov 19 at 12:30
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
Nov 19 at 13:16
Does this answer the question?
– Szeto
Nov 19 at 12:26
Does this answer the question?
– Szeto
Nov 19 at 12:26
1
1
Well it does if you are tasked with doing the integral from $0$ to $infty$ and mistakenly assume that the first thing to do must be to figure out the indefinite integral, and then stick the limits in.
– Richard Martin
Nov 19 at 12:30
Well it does if you are tasked with doing the integral from $0$ to $infty$ and mistakenly assume that the first thing to do must be to figure out the indefinite integral, and then stick the limits in.
– Richard Martin
Nov 19 at 12:30
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
Nov 19 at 13:16
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
Nov 19 at 13:16
add a comment |
up vote
-1
down vote
You can also do this:
When you have factorised the term $x^4+1$ into factors $p(x) & q(x)$ then
$$ frac{1}{x^4+1} = frac {Ax+M}{p(x)}+frac{Bx+N}{q(x)}$$ and then to find A, B, M and N just take values of x for both sides and equate to find them just like linear equations in two variables.
answered Nov 19 at 9:45
jayant98
35414
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
Nov 19 at 9:46
One of easiest value putting is x=0 and if you can see x=√2 also helps! Also if you see the IIT-JEE papers you will find these types of problem to be solved in less than 5 minutes and the method taught to the aspirants is the same method. It is just the issue of finding so called special x value so that equation is simpler. And for that my friend practice is the best method.
– jayant98
Nov 19 at 9:52
I have tried those values and they do not solve my problem also
– hopefully
Nov 19 at 9:56
$x=0$ implies $1=M+N$, $x = sqrt(2)$ implies $1 = Ax+M + 5Bx+5N$, using $-sqrt(2)$ analogue $1 = -5Ax+5M -Bx+N$, i.e. $Ax+5Bx = -5Ax-Bx$, i.e. $A+B=0$. Now get a handy 4th equation ...
– Stockfish
Nov 19 at 10:49
add a comment |
up vote
-1
down vote
You can also do this:
When you have factorised the term $x^4+1$ into factors $p(x) & q(x)$ then
$$ frac{1}{x^4+1} = frac {Ax+M}{p(x)}+frac{Bx+N}{q(x)}$$ and then to find A, B, M and N just take values of x for both sides and equate to find them just like linear equations in two variables.
answered Nov 19 at 9:45
jayant98
35414
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
Nov 19 at 9:46
One of easiest value putting is x=0 and if you can see x=√2 also helps! Also if you see the IIT-JEE papers you will find these types of problem to be solved in less than 5 minutes and the method taught to the aspirants is the same method. It is just the issue of finding so called special x value so that equation is simpler. And for that my friend practice is the best method.
– jayant98
Nov 19 at 9:52
I have tried those values and they do not solve my problem also
– hopefully
Nov 19 at 9:56
$x=0$ implies $1=M+N$, $x = sqrt(2)$ implies $1 = Ax+M + 5Bx+5N$, using $-sqrt(2)$ analogue $1 = -5Ax+5M -Bx+N$, i.e. $Ax+5Bx = -5Ax-Bx$, i.e. $A+B=0$. Now get a handy 4th equation ...
– Stockfish
Nov 19 at 10:49
add a comment |
up vote
-1
down vote
up vote
-1
down vote
You can also do this:
When you have factorised the term $x^4+1$ into factors $p(x) & q(x)$ then
$$ frac{1}{x^4+1} = frac {Ax+M}{p(x)}+frac{Bx+N}{q(x)}$$ and then to find A, B, M and N just take values of x for both sides and equate to find them just like linear equations in two variables.
answered Nov 19 at 9:45
jayant98
35414
You can also do this:
When you have factorised the term $x^4+1$ into factors $p(x) & q(x)$ then
$$ frac{1}{x^4+1} = frac {Ax+M}{p(x)}+frac{Bx+N}{q(x)}$$ and then to find A, B, M and N just take values of x for both sides and equate to find them just like linear equations in two variables.
answered Nov 19 at 9:45
jayant98
35414
edited Nov 19 at 10:05
answered Nov 19 at 9:45
jayant98
35414
answered Nov 19 at 9:45
jayant98
35414
answered Nov 19 at 9:45
jayant98
35414
35414
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
Nov 19 at 9:46
One of easiest value putting is x=0 and if you can see x=√2 also helps! Also if you see the IIT-JEE papers you will find these types of problem to be solved in less than 5 minutes and the method taught to the aspirants is the same method. It is just the issue of finding so called special x value so that equation is simpler. And for that my friend practice is the best method.
– jayant98
Nov 19 at 9:52
I have tried those values and they do not solve my problem also
– hopefully
Nov 19 at 9:56
$x=0$ implies $1=M+N$, $x = sqrt(2)$ implies $1 = Ax+M + 5Bx+5N$, using $-sqrt(2)$ analogue $1 = -5Ax+5M -Bx+N$, i.e. $Ax+5Bx = -5Ax-Bx$, i.e. $A+B=0$. Now get a handy 4th equation ...
– Stockfish
Nov 19 at 10:49
add a comment |
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
Nov 19 at 9:46
One of easiest value putting is x=0 and if you can see x=√2 also helps! Also if you see the IIT-JEE papers you will find these types of problem to be solved in less than 5 minutes and the method taught to the aspirants is the same method. It is just the issue of finding so called special x value so that equation is simpler. And for that my friend practice is the best method.
– jayant98
Nov 19 at 9:52
I have tried those values and they do not solve my problem also
– hopefully
Nov 19 at 9:56
$x=0$ implies $1=M+N$, $x = sqrt(2)$ implies $1 = Ax+M + 5Bx+5N$, using $-sqrt(2)$ analogue $1 = -5Ax+5M -Bx+N$, i.e. $Ax+5Bx = -5Ax-Bx$, i.e. $A+B=0$. Now get a handy 4th equation ...
– Stockfish
Nov 19 at 10:49
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
Nov 19 at 9:46
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
Nov 19 at 9:46
One of easiest value putting is x=0 and if you can see x=√2 also helps! Also if you see the IIT-JEE papers you will find these types of problem to be solved in less than 5 minutes and the method taught to the aspirants is the same method. It is just the issue of finding so called special x value so that equation is simpler. And for that my friend practice is the best method.
– jayant98
Nov 19 at 9:52
One of easiest value putting is x=0 and if you can see x=√2 also helps! Also if you see the IIT-JEE papers you will find these types of problem to be solved in less than 5 minutes and the method taught to the aspirants is the same method. It is just the issue of finding so called special x value so that equation is simpler. And for that my friend practice is the best method.
– jayant98
Nov 19 at 9:52
I have tried those values and they do not solve my problem also
– hopefully
Nov 19 at 9:56
I have tried those values and they do not solve my problem also
– hopefully
Nov 19 at 9:56
$x=0$ implies $1=M+N$, $x = sqrt(2)$ implies $1 = Ax+M + 5Bx+5N$, using $-sqrt(2)$ analogue $1 = -5Ax+5M -Bx+N$, i.e. $Ax+5Bx = -5Ax-Bx$, i.e. $A+B=0$. Now get a handy 4th equation ...
– Stockfish
Nov 19 at 10:49
$x=0$ implies $1=M+N$, $x = sqrt(2)$ implies $1 = Ax+M + 5Bx+5N$, using $-sqrt(2)$ analogue $1 = -5Ax+5M -Bx+N$, i.e. $Ax+5Bx = -5Ax-Bx$, i.e. $A+B=0$. Now get a handy 4th equation ...
– Stockfish
Nov 19 at 10:49
add a comment |
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Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
Nov 19 at 10:16
Evaluation of integral of $frac{1}{x^4 + 1}$ AND evaluation of integral of $frac{1}{x^5 + 1}$ AND evaluation of integral of $frac{1}{x^6 + 1}$. See this also.
– Dave L. Renfro
Nov 19 at 13:55