Mixing
Evaluate the integral $int frac {1}{x^4 +1} dx$( finding the constants in the partial fraction method).
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0
down vote
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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 9 hours ago
hopefully
2079
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 9 hours ago
hopefully
2079
2
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
8 hours ago
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
5 hours ago
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 9 hours ago
hopefully
2079
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 9 hours ago
hopefully
2079
asked 9 hours ago
hopefully
2079
asked 9 hours ago
hopefully
2079
asked 9 hours ago
hopefully
2079
asked 9 hours ago
hopefully
2079
2079
2
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
8 hours ago
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
5 hours ago
add a comment |
2
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
8 hours ago
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
5 hours ago
2
2
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
8 hours ago
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
8 hours ago
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
5 hours ago
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
5 hours ago
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 8 hours ago
Richard Martin
1,2588
Does this answer the question?
– Szeto
6 hours ago
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
6 hours ago
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
5 hours ago
add a comment |
up vote
0
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 9 hours ago
jayant98
12613
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
9 hours ago
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
9 hours ago
I have tried those values and they do not solve my problem also
– hopefully
8 hours ago
$x=0$ 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
8 hours ago
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 8 hours ago
Richard Martin
1,2588
Does this answer the question?
– Szeto
6 hours ago
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
6 hours ago
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
5 hours ago
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 8 hours ago
Richard Martin
1,2588
Does this answer the question?
– Szeto
6 hours ago
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
6 hours ago
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
5 hours ago
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 8 hours ago
Richard Martin
1,2588
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 8 hours ago
Richard Martin
1,2588
edited 8 hours ago
answered 8 hours ago
Richard Martin
1,2588
answered 8 hours ago
Richard Martin
1,2588
answered 8 hours ago
Richard Martin
1,2588
1,2588
Does this answer the question?
– Szeto
6 hours ago
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
6 hours ago
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
5 hours ago
add a comment |
Does this answer the question?
– Szeto
6 hours ago
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
6 hours ago
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
5 hours ago
Does this answer the question?
– Szeto
6 hours ago
Does this answer the question?
– Szeto
6 hours ago
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
6 hours ago
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
6 hours ago
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
5 hours ago
Which, I hasten to add, is exactly what I would have done when I was twelve!
– Richard Martin
5 hours ago
add a comment |
up vote
0
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 9 hours ago
jayant98
12613
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
9 hours ago
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
9 hours ago
I have tried those values and they do not solve my problem also
– hopefully
8 hours ago
$x=0$ 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
8 hours ago
add a comment |
up vote
0
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 9 hours ago
jayant98
12613
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
9 hours ago
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
9 hours ago
I have tried those values and they do not solve my problem also
– hopefully
8 hours ago
$x=0$ 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
8 hours ago
add a comment |
up vote
0
down vote
up vote
0
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 9 hours ago
jayant98
12613
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 9 hours ago
jayant98
12613
edited 8 hours ago
answered 9 hours ago
jayant98
12613
answered 9 hours ago
jayant98
12613
answered 9 hours ago
jayant98
12613
12613
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
9 hours ago
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
9 hours ago
I have tried those values and they do not solve my problem also
– hopefully
8 hours ago
$x=0$ 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
8 hours ago
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
9 hours ago
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
9 hours ago
I have tried those values and they do not solve my problem also
– hopefully
8 hours ago
$x=0$ 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
8 hours ago
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
9 hours ago
I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult
– hopefully
9 hours ago
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
9 hours ago
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
9 hours ago
I have tried those values and they do not solve my problem also
– hopefully
8 hours ago
I have tried those values and they do not solve my problem also
– hopefully
8 hours ago
$x=0$ 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
8 hours ago
$x=0$ 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
8 hours ago
add a comment |
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2
Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(sqrt{2}x)^2$.
– Batominovski
8 hours ago
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
5 hours ago