N-dependent (even) function integral












5














We want to compute, for any $n in mathbb{N}$ the following integral: $$int_{-1}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx$$



My attempt:
if $n$ is odd, the integral is trivially equal to $0$ since $x^n/(sqrt[n]{1+x}+sqrt[n]{1-x})$ is itself an odd function. If $n$ is even, so does $x^n/(sqrt[n]{1+x}+sqrt[n]{1-x})$ and therefore $$int_{-1}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx=2int_{0}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx$$ but this is not really helping. Any ideas?
Thanks






calculus integration analysis definite-integrals even-and-odd-functions







share|cite|improve this question




















  • 2




    Is there any particular reason you expect the integral to have a closed form?
    – 5xum
    Aug 10 at 11:21






  • 1




    I would start by forcing a factor of $(1+x)^{1/n}$ out of the denominator and substitute $frac{1-x}{1+x}=t$
    – Zacky
    Aug 10 at 11:23






  • 1




    For shortness, $sqrt[n]{1+x}=a$, $sqrt[n]{1-x}=b$. Multiply above and below by $a^{n-1}b^0-a^{n-2}b^1+cdots+a^0 b^{n-1}$ so you no longer have a fraction but a sum. However, it may not help much, you still have to integrate integrals of the form $int x^n sqrt[n]{(1-x)^k (1+x)^{n-1-k}}dx$.
    – orion
    Aug 10 at 11:42












  • Maybe this identity comes in handy: $$ int_0^1 x^{a-1}(1-x)^{b-1} frac{dx}{(x+p)^{a+b}} = frac{Gamma(a)Gamma(b)}{Gamma(a+b)}frac{1}{(1+p)^{a} p^b}$$
    – Zacky
    Aug 10 at 11:47






  • 1




    Just by checking Mathematica output, it doesn't look like there's a nice closed form solution in general.
    – orion
    Aug 10 at 12:04
















5














We want to compute, for any $n in mathbb{N}$ the following integral: $$int_{-1}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx$$



My attempt:
if $n$ is odd, the integral is trivially equal to $0$ since $x^n/(sqrt[n]{1+x}+sqrt[n]{1-x})$ is itself an odd function. If $n$ is even, so does $x^n/(sqrt[n]{1+x}+sqrt[n]{1-x})$ and therefore $$int_{-1}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx=2int_{0}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx$$ but this is not really helping. Any ideas?
Thanks






calculus integration analysis definite-integrals even-and-odd-functions







share|cite|improve this question




















  • 2




    Is there any particular reason you expect the integral to have a closed form?
    – 5xum
    Aug 10 at 11:21






  • 1




    I would start by forcing a factor of $(1+x)^{1/n}$ out of the denominator and substitute $frac{1-x}{1+x}=t$
    – Zacky
    Aug 10 at 11:23






  • 1




    For shortness, $sqrt[n]{1+x}=a$, $sqrt[n]{1-x}=b$. Multiply above and below by $a^{n-1}b^0-a^{n-2}b^1+cdots+a^0 b^{n-1}$ so you no longer have a fraction but a sum. However, it may not help much, you still have to integrate integrals of the form $int x^n sqrt[n]{(1-x)^k (1+x)^{n-1-k}}dx$.
    – orion
    Aug 10 at 11:42












  • Maybe this identity comes in handy: $$ int_0^1 x^{a-1}(1-x)^{b-1} frac{dx}{(x+p)^{a+b}} = frac{Gamma(a)Gamma(b)}{Gamma(a+b)}frac{1}{(1+p)^{a} p^b}$$
    – Zacky
    Aug 10 at 11:47






  • 1




    Just by checking Mathematica output, it doesn't look like there's a nice closed form solution in general.
    – orion
    Aug 10 at 12:04














5












5








5


1





We want to compute, for any $n in mathbb{N}$ the following integral: $$int_{-1}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx$$



My attempt:
if $n$ is odd, the integral is trivially equal to $0$ since $x^n/(sqrt[n]{1+x}+sqrt[n]{1-x})$ is itself an odd function. If $n$ is even, so does $x^n/(sqrt[n]{1+x}+sqrt[n]{1-x})$ and therefore $$int_{-1}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx=2int_{0}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx$$ but this is not really helping. Any ideas?
Thanks






calculus integration analysis definite-integrals even-and-odd-functions







share|cite|improve this question















We want to compute, for any $n in mathbb{N}$ the following integral: $$int_{-1}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx$$



My attempt:
if $n$ is odd, the integral is trivially equal to $0$ since $x^n/(sqrt[n]{1+x}+sqrt[n]{1-x})$ is itself an odd function. If $n$ is even, so does $x^n/(sqrt[n]{1+x}+sqrt[n]{1-x})$ and therefore $$int_{-1}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx=2int_{0}^{1} frac{x^n}{sqrt[n]{1+x}+sqrt[n]{1-x}}dx$$ but this is not really helping. Any ideas?
Thanks






calculus integration analysis definite-integrals even-and-odd-functions




calculus integration analysis definite-integrals even-and-odd-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 6:37









Robert Howard

1,9161822




1,9161822










asked Aug 10 at 11:05









arnett

262




262








  • 2




    Is there any particular reason you expect the integral to have a closed form?
    – 5xum
    Aug 10 at 11:21






  • 1




    I would start by forcing a factor of $(1+x)^{1/n}$ out of the denominator and substitute $frac{1-x}{1+x}=t$
    – Zacky
    Aug 10 at 11:23






  • 1




    For shortness, $sqrt[n]{1+x}=a$, $sqrt[n]{1-x}=b$. Multiply above and below by $a^{n-1}b^0-a^{n-2}b^1+cdots+a^0 b^{n-1}$ so you no longer have a fraction but a sum. However, it may not help much, you still have to integrate integrals of the form $int x^n sqrt[n]{(1-x)^k (1+x)^{n-1-k}}dx$.
    – orion
    Aug 10 at 11:42












  • Maybe this identity comes in handy: $$ int_0^1 x^{a-1}(1-x)^{b-1} frac{dx}{(x+p)^{a+b}} = frac{Gamma(a)Gamma(b)}{Gamma(a+b)}frac{1}{(1+p)^{a} p^b}$$
    – Zacky
    Aug 10 at 11:47






  • 1




    Just by checking Mathematica output, it doesn't look like there's a nice closed form solution in general.
    – orion
    Aug 10 at 12:04














  • 2




    Is there any particular reason you expect the integral to have a closed form?
    – 5xum
    Aug 10 at 11:21






  • 1




    I would start by forcing a factor of $(1+x)^{1/n}$ out of the denominator and substitute $frac{1-x}{1+x}=t$
    – Zacky
    Aug 10 at 11:23






  • 1




    For shortness, $sqrt[n]{1+x}=a$, $sqrt[n]{1-x}=b$. Multiply above and below by $a^{n-1}b^0-a^{n-2}b^1+cdots+a^0 b^{n-1}$ so you no longer have a fraction but a sum. However, it may not help much, you still have to integrate integrals of the form $int x^n sqrt[n]{(1-x)^k (1+x)^{n-1-k}}dx$.
    – orion
    Aug 10 at 11:42












  • Maybe this identity comes in handy: $$ int_0^1 x^{a-1}(1-x)^{b-1} frac{dx}{(x+p)^{a+b}} = frac{Gamma(a)Gamma(b)}{Gamma(a+b)}frac{1}{(1+p)^{a} p^b}$$
    – Zacky
    Aug 10 at 11:47






  • 1




    Just by checking Mathematica output, it doesn't look like there's a nice closed form solution in general.
    – orion
    Aug 10 at 12:04








2




2




Is there any particular reason you expect the integral to have a closed form?
– 5xum
Aug 10 at 11:21




Is there any particular reason you expect the integral to have a closed form?
– 5xum
Aug 10 at 11:21




1




1




I would start by forcing a factor of $(1+x)^{1/n}$ out of the denominator and substitute $frac{1-x}{1+x}=t$
– Zacky
Aug 10 at 11:23




I would start by forcing a factor of $(1+x)^{1/n}$ out of the denominator and substitute $frac{1-x}{1+x}=t$
– Zacky
Aug 10 at 11:23




1




1




For shortness, $sqrt[n]{1+x}=a$, $sqrt[n]{1-x}=b$. Multiply above and below by $a^{n-1}b^0-a^{n-2}b^1+cdots+a^0 b^{n-1}$ so you no longer have a fraction but a sum. However, it may not help much, you still have to integrate integrals of the form $int x^n sqrt[n]{(1-x)^k (1+x)^{n-1-k}}dx$.
– orion
Aug 10 at 11:42






For shortness, $sqrt[n]{1+x}=a$, $sqrt[n]{1-x}=b$. Multiply above and below by $a^{n-1}b^0-a^{n-2}b^1+cdots+a^0 b^{n-1}$ so you no longer have a fraction but a sum. However, it may not help much, you still have to integrate integrals of the form $int x^n sqrt[n]{(1-x)^k (1+x)^{n-1-k}}dx$.
– orion
Aug 10 at 11:42














Maybe this identity comes in handy: $$ int_0^1 x^{a-1}(1-x)^{b-1} frac{dx}{(x+p)^{a+b}} = frac{Gamma(a)Gamma(b)}{Gamma(a+b)}frac{1}{(1+p)^{a} p^b}$$
– Zacky
Aug 10 at 11:47




Maybe this identity comes in handy: $$ int_0^1 x^{a-1}(1-x)^{b-1} frac{dx}{(x+p)^{a+b}} = frac{Gamma(a)Gamma(b)}{Gamma(a+b)}frac{1}{(1+p)^{a} p^b}$$
– Zacky
Aug 10 at 11:47




1




1




Just by checking Mathematica output, it doesn't look like there's a nice closed form solution in general.
– orion
Aug 10 at 12:04




Just by checking Mathematica output, it doesn't look like there's a nice closed form solution in general.
– orion
Aug 10 at 12:04















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