Does $int_{-1}^1 frac{f(x)g(x)}{sqrt(1-x^2)} dx$ converge?
I was supposed to prove that $int_{-1}^1 frac{f(x)g(x)}{sqrt(1-x^2)} dx$ converges.
However, I am not sure if this actually converges because I was not given any information about $f(x)$ and $g(x)$ other than that $f(x), g(x) in mathbb{R}[x]$. Could someone give any insights? Thanks
calculus integration convergence improper-integrals
asked Nov 30 at 6:01
dmsj djsl
35517
add a comment |
I was supposed to prove that $int_{-1}^1 frac{f(x)g(x)}{sqrt(1-x^2)} dx$ converges.
However, I am not sure if this actually converges because I was not given any information about $f(x)$ and $g(x)$ other than that $f(x), g(x) in mathbb{R}[x]$. Could someone give any insights? Thanks
calculus integration convergence improper-integrals
asked Nov 30 at 6:01
dmsj djsl
35517
1
A continuous function on a compact set is bounded. Then you are left to prove that $int_{-1}^1 frac{1}{sqrt{1-x^2}} dx$ converges
– nicomezi
Nov 30 at 6:08
@dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
– Robert Z
Nov 30 at 6:18
add a comment |
I was supposed to prove that $int_{-1}^1 frac{f(x)g(x)}{sqrt(1-x^2)} dx$ converges.
However, I am not sure if this actually converges because I was not given any information about $f(x)$ and $g(x)$ other than that $f(x), g(x) in mathbb{R}[x]$. Could someone give any insights? Thanks
calculus integration convergence improper-integrals
asked Nov 30 at 6:01
dmsj djsl
35517
I was supposed to prove that $int_{-1}^1 frac{f(x)g(x)}{sqrt(1-x^2)} dx$ converges.
However, I am not sure if this actually converges because I was not given any information about $f(x)$ and $g(x)$ other than that $f(x), g(x) in mathbb{R}[x]$. Could someone give any insights? Thanks
calculus integration convergence improper-integrals
calculus integration convergence improper-integrals
asked Nov 30 at 6:01
dmsj djsl
35517
asked Nov 30 at 6:01
dmsj djsl
35517
asked Nov 30 at 6:01
dmsj djsl
35517
asked Nov 30 at 6:01
dmsj djsl
35517
asked Nov 30 at 6:01
dmsj djsl
35517
35517
1
A continuous function on a compact set is bounded. Then you are left to prove that $int_{-1}^1 frac{1}{sqrt{1-x^2}} dx$ converges
– nicomezi
Nov 30 at 6:08
@dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
– Robert Z
Nov 30 at 6:18
add a comment |
1
A continuous function on a compact set is bounded. Then you are left to prove that $int_{-1}^1 frac{1}{sqrt{1-x^2}} dx$ converges
– nicomezi
Nov 30 at 6:08
@dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
– Robert Z
Nov 30 at 6:18
1
1
A continuous function on a compact set is bounded. Then you are left to prove that $int_{-1}^1 frac{1}{sqrt{1-x^2}} dx$ converges
– nicomezi
Nov 30 at 6:08
A continuous function on a compact set is bounded. Then you are left to prove that $int_{-1}^1 frac{1}{sqrt{1-x^2}} dx$ converges
– nicomezi
Nov 30 at 6:08
@dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
– Robert Z
Nov 30 at 6:18
@dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
– Robert Z
Nov 30 at 6:18
add a comment |
1 Answer
1
active
oldest
votes
Are $f$ and $g$ continuous? If they are, then the only possibility of non-convergence is at the endpoints $pm1$ of the interval. So we ask if
the integrals over $[1-epsilon,1]$ and $[-1,-1+epsilon]$ converge. These
are similar, so look at the second. A change of variables gives
$$int_{-1}^{-1+epsilon}frac{h(x),dx}{sqrt{1-x^2}}
=int_0^epsilonfrac{h(y-1),dy}{sqrt{y(2-y)}}=int_0^epsilonfrac
{k(y),dy}{sqrt y}$$
for suitable continuous functions $h$ and $k$. As $k$ is continuous,
it is bounded on $[0,epsilon]$, so the integral is bounded by a constant
multiple of $int_0^epsilon frac{dy}{sqrt y}$ which is a convergent
integral.
answered Nov 30 at 6:13
Lord Shark the Unknown
101k958131
In fact, $f,g in Bbb R[x]$ so they are continuous.
– Guacho Perez
Nov 30 at 6:16
add a comment |
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Are $f$ and $g$ continuous? If they are, then the only possibility of non-convergence is at the endpoints $pm1$ of the interval. So we ask if
the integrals over $[1-epsilon,1]$ and $[-1,-1+epsilon]$ converge. These
are similar, so look at the second. A change of variables gives
$$int_{-1}^{-1+epsilon}frac{h(x),dx}{sqrt{1-x^2}}
=int_0^epsilonfrac{h(y-1),dy}{sqrt{y(2-y)}}=int_0^epsilonfrac
{k(y),dy}{sqrt y}$$
for suitable continuous functions $h$ and $k$. As $k$ is continuous,
it is bounded on $[0,epsilon]$, so the integral is bounded by a constant
multiple of $int_0^epsilon frac{dy}{sqrt y}$ which is a convergent
integral.
answered Nov 30 at 6:13
Lord Shark the Unknown
101k958131
In fact, $f,g in Bbb R[x]$ so they are continuous.
– Guacho Perez
Nov 30 at 6:16
add a comment |
Are $f$ and $g$ continuous? If they are, then the only possibility of non-convergence is at the endpoints $pm1$ of the interval. So we ask if
the integrals over $[1-epsilon,1]$ and $[-1,-1+epsilon]$ converge. These
are similar, so look at the second. A change of variables gives
$$int_{-1}^{-1+epsilon}frac{h(x),dx}{sqrt{1-x^2}}
=int_0^epsilonfrac{h(y-1),dy}{sqrt{y(2-y)}}=int_0^epsilonfrac
{k(y),dy}{sqrt y}$$
for suitable continuous functions $h$ and $k$. As $k$ is continuous,
it is bounded on $[0,epsilon]$, so the integral is bounded by a constant
multiple of $int_0^epsilon frac{dy}{sqrt y}$ which is a convergent
integral.
answered Nov 30 at 6:13
Lord Shark the Unknown
101k958131
In fact, $f,g in Bbb R[x]$ so they are continuous.
– Guacho Perez
Nov 30 at 6:16
add a comment |
Are $f$ and $g$ continuous? If they are, then the only possibility of non-convergence is at the endpoints $pm1$ of the interval. So we ask if
the integrals over $[1-epsilon,1]$ and $[-1,-1+epsilon]$ converge. These
are similar, so look at the second. A change of variables gives
$$int_{-1}^{-1+epsilon}frac{h(x),dx}{sqrt{1-x^2}}
=int_0^epsilonfrac{h(y-1),dy}{sqrt{y(2-y)}}=int_0^epsilonfrac
{k(y),dy}{sqrt y}$$
for suitable continuous functions $h$ and $k$. As $k$ is continuous,
it is bounded on $[0,epsilon]$, so the integral is bounded by a constant
multiple of $int_0^epsilon frac{dy}{sqrt y}$ which is a convergent
integral.
answered Nov 30 at 6:13
Lord Shark the Unknown
101k958131
Are $f$ and $g$ continuous? If they are, then the only possibility of non-convergence is at the endpoints $pm1$ of the interval. So we ask if
the integrals over $[1-epsilon,1]$ and $[-1,-1+epsilon]$ converge. These
are similar, so look at the second. A change of variables gives
$$int_{-1}^{-1+epsilon}frac{h(x),dx}{sqrt{1-x^2}}
=int_0^epsilonfrac{h(y-1),dy}{sqrt{y(2-y)}}=int_0^epsilonfrac
{k(y),dy}{sqrt y}$$
for suitable continuous functions $h$ and $k$. As $k$ is continuous,
it is bounded on $[0,epsilon]$, so the integral is bounded by a constant
multiple of $int_0^epsilon frac{dy}{sqrt y}$ which is a convergent
integral.
answered Nov 30 at 6:13
Lord Shark the Unknown
101k958131
answered Nov 30 at 6:13
Lord Shark the Unknown
101k958131
answered Nov 30 at 6:13
Lord Shark the Unknown
101k958131
answered Nov 30 at 6:13
Lord Shark the Unknown
101k958131
101k958131
In fact, $f,g in Bbb R[x]$ so they are continuous.
– Guacho Perez
Nov 30 at 6:16
add a comment |
In fact, $f,g in Bbb R[x]$ so they are continuous.
– Guacho Perez
Nov 30 at 6:16
In fact, $f,g in Bbb R[x]$ so they are continuous.
– Guacho Perez
Nov 30 at 6:16
In fact, $f,g in Bbb R[x]$ so they are continuous.
– Guacho Perez
Nov 30 at 6:16
add a comment |
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1
A continuous function on a compact set is bounded. Then you are left to prove that $int_{-1}^1 frac{1}{sqrt{1-x^2}} dx$ converges
– nicomezi
Nov 30 at 6:08
@dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
– Robert Z
Nov 30 at 6:18