Does $int_{-1}^1 frac{f(x)g(x)}{sqrt(1-x^2)} dx$ converge?












0












$begingroup$


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







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  • 1




    $begingroup$
    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
    $endgroup$
    – nicomezi
    Nov 30 '18 at 6:08












  • $begingroup$
    @dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
    $endgroup$
    – Robert Z
    Nov 30 '18 at 6:18


















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    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
    $endgroup$
    – nicomezi
    Nov 30 '18 at 6:08












  • $begingroup$
    @dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
    $endgroup$
    – Robert Z
    Nov 30 '18 at 6:18
















0












0








0


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






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share|cite|improve this question











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share|cite|improve this question










asked Nov 30 '18 at 6:01









dmsj djsldmsj djsl

35517




35517








  • 1




    $begingroup$
    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
    $endgroup$
    – nicomezi
    Nov 30 '18 at 6:08












  • $begingroup$
    @dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
    $endgroup$
    – Robert Z
    Nov 30 '18 at 6:18
















  • 1




    $begingroup$
    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
    $endgroup$
    – nicomezi
    Nov 30 '18 at 6:08












  • $begingroup$
    @dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
    $endgroup$
    – Robert Z
    Nov 30 '18 at 6:18










1




1




$begingroup$
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
$endgroup$
– nicomezi
Nov 30 '18 at 6:08






$begingroup$
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
$endgroup$
– nicomezi
Nov 30 '18 at 6:08














$begingroup$
@dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
$endgroup$
– Robert Z
Nov 30 '18 at 6:18






$begingroup$
@dmsj So $f$ and $g$ are real polynomials ($mathbb{R}[x]$). Aren't they?
$endgroup$
– Robert Z
Nov 30 '18 at 6:18












1 Answer
1






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact, $f,g in Bbb R[x]$ so they are continuous.
    $endgroup$
    – Guacho Perez
    Nov 30 '18 at 6:16











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1 Answer
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1 Answer
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active

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2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact, $f,g in Bbb R[x]$ so they are continuous.
    $endgroup$
    – Guacho Perez
    Nov 30 '18 at 6:16
















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact, $f,g in Bbb R[x]$ so they are continuous.
    $endgroup$
    – Guacho Perez
    Nov 30 '18 at 6:16














2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 '18 at 6:13









Lord Shark the UnknownLord Shark the Unknown

104k1160132




104k1160132












  • $begingroup$
    In fact, $f,g in Bbb R[x]$ so they are continuous.
    $endgroup$
    – Guacho Perez
    Nov 30 '18 at 6:16


















  • $begingroup$
    In fact, $f,g in Bbb R[x]$ so they are continuous.
    $endgroup$
    – Guacho Perez
    Nov 30 '18 at 6:16
















$begingroup$
In fact, $f,g in Bbb R[x]$ so they are continuous.
$endgroup$
– Guacho Perez
Nov 30 '18 at 6:16




$begingroup$
In fact, $f,g in Bbb R[x]$ so they are continuous.
$endgroup$
– Guacho Perez
Nov 30 '18 at 6:16


















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