Integral of Modified Bessel Function

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Evaluate:
$$intlimits_x=0^infty e^-ax^2I_0(x) log_2(I_0(x))x , dx$$



where $I_0(x)$ is the modified Bessel function of first kind and zeroth order, and $alpha>0$.




I can find upper bounds and lower bounds using the expansion of $I_0(x)$; however, they are not enough. Any idea how to solve this?




integration definite-integrals bessel-functions




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  • 1




    A not-so-terrible upper bound can be derived from $log I_0(x)leq x$. We have $$int_0^+infty xcdot I_0(x) e^-ax^2,dx = frac12ae^frac14a.$$
    – Jack D'Aurizio♦
    Jul 25 at 20:39










  • @JackD'Aurizio Thanks. The diamond means you are one of the moderators?
    – Susan_Math123
    Jul 25 at 21:05










  • You're welcome, and... yup.
    – Jack D'Aurizio♦
    Jul 25 at 21:16










  • @JackD'Aurizio To get another bound I can use $ln(x)leq x-1$. Then I will need to evaluate $int_0^inftyx I_0(x)^2 e^-a x^2 dx$. Do you have an answer for this as well?
    – Susan_Math123
    Jul 26 at 16:30










  • Such bound is much worse than the previous one, but in any case $$int_0^+inftyx I_0(x)^2 e^-ax^2,dx = frac12a e^1/(2a) I_0(1/(2a)).$$
    – Jack D'Aurizio♦
    Jul 26 at 16:42














up vote
2
down vote

favorite
1













Evaluate:
$$intlimits_x=0^infty e^-ax^2I_0(x) log_2(I_0(x))x , dx$$



where $I_0(x)$ is the modified Bessel function of first kind and zeroth order, and $alpha>0$.




I can find upper bounds and lower bounds using the expansion of $I_0(x)$; however, they are not enough. Any idea how to solve this?




integration definite-integrals bessel-functions




share|cite|improve this question

















  • 1




    A not-so-terrible upper bound can be derived from $log I_0(x)leq x$. We have $$int_0^+infty xcdot I_0(x) e^-ax^2,dx = frac12ae^frac14a.$$
    – Jack D'Aurizio♦
    Jul 25 at 20:39










  • @JackD'Aurizio Thanks. The diamond means you are one of the moderators?
    – Susan_Math123
    Jul 25 at 21:05










  • You're welcome, and... yup.
    – Jack D'Aurizio♦
    Jul 25 at 21:16










  • @JackD'Aurizio To get another bound I can use $ln(x)leq x-1$. Then I will need to evaluate $int_0^inftyx I_0(x)^2 e^-a x^2 dx$. Do you have an answer for this as well?
    – Susan_Math123
    Jul 26 at 16:30










  • Such bound is much worse than the previous one, but in any case $$int_0^+inftyx I_0(x)^2 e^-ax^2,dx = frac12a e^1/(2a) I_0(1/(2a)).$$
    – Jack D'Aurizio♦
    Jul 26 at 16:42












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1






Evaluate:
$$intlimits_x=0^infty e^-ax^2I_0(x) log_2(I_0(x))x , dx$$



where $I_0(x)$ is the modified Bessel function of first kind and zeroth order, and $alpha>0$.




I can find upper bounds and lower bounds using the expansion of $I_0(x)$; however, they are not enough. Any idea how to solve this?




integration definite-integrals bessel-functions




share|cite|improve this question














Evaluate:
$$intlimits_x=0^infty e^-ax^2I_0(x) log_2(I_0(x))x , dx$$



where $I_0(x)$ is the modified Bessel function of first kind and zeroth order, and $alpha>0$.




I can find upper bounds and lower bounds using the expansion of $I_0(x)$; however, they are not enough. Any idea how to solve this?





integration definite-integrals bessel-functions





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 17:02









Michael Hardy

204k23186461




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asked Jul 25 at 19:52









Susan_Math123

825319




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  • 1




    A not-so-terrible upper bound can be derived from $log I_0(x)leq x$. We have $$int_0^+infty xcdot I_0(x) e^-ax^2,dx = frac12ae^frac14a.$$
    – Jack D'Aurizio♦
    Jul 25 at 20:39










  • @JackD'Aurizio Thanks. The diamond means you are one of the moderators?
    – Susan_Math123
    Jul 25 at 21:05










  • You're welcome, and... yup.
    – Jack D'Aurizio♦
    Jul 25 at 21:16










  • @JackD'Aurizio To get another bound I can use $ln(x)leq x-1$. Then I will need to evaluate $int_0^inftyx I_0(x)^2 e^-a x^2 dx$. Do you have an answer for this as well?
    – Susan_Math123
    Jul 26 at 16:30










  • Such bound is much worse than the previous one, but in any case $$int_0^+inftyx I_0(x)^2 e^-ax^2,dx = frac12a e^1/(2a) I_0(1/(2a)).$$
    – Jack D'Aurizio♦
    Jul 26 at 16:42












  • 1




    A not-so-terrible upper bound can be derived from $log I_0(x)leq x$. We have $$int_0^+infty xcdot I_0(x) e^-ax^2,dx = frac12ae^frac14a.$$
    – Jack D'Aurizio♦
    Jul 25 at 20:39










  • @JackD'Aurizio Thanks. The diamond means you are one of the moderators?
    – Susan_Math123
    Jul 25 at 21:05










  • You're welcome, and... yup.
    – Jack D'Aurizio♦
    Jul 25 at 21:16










  • @JackD'Aurizio To get another bound I can use $ln(x)leq x-1$. Then I will need to evaluate $int_0^inftyx I_0(x)^2 e^-a x^2 dx$. Do you have an answer for this as well?
    – Susan_Math123
    Jul 26 at 16:30










  • Such bound is much worse than the previous one, but in any case $$int_0^+inftyx I_0(x)^2 e^-ax^2,dx = frac12a e^1/(2a) I_0(1/(2a)).$$
    – Jack D'Aurizio♦
    Jul 26 at 16:42







1




1




A not-so-terrible upper bound can be derived from $log I_0(x)leq x$. We have $$int_0^+infty xcdot I_0(x) e^-ax^2,dx = frac12ae^frac14a.$$
– Jack D'Aurizio♦
Jul 25 at 20:39




A not-so-terrible upper bound can be derived from $log I_0(x)leq x$. We have $$int_0^+infty xcdot I_0(x) e^-ax^2,dx = frac12ae^frac14a.$$
– Jack D'Aurizio♦
Jul 25 at 20:39












@JackD'Aurizio Thanks. The diamond means you are one of the moderators?
– Susan_Math123
Jul 25 at 21:05




@JackD'Aurizio Thanks. The diamond means you are one of the moderators?
– Susan_Math123
Jul 25 at 21:05












You're welcome, and... yup.
– Jack D'Aurizio♦
Jul 25 at 21:16




You're welcome, and... yup.
– Jack D'Aurizio♦
Jul 25 at 21:16












@JackD'Aurizio To get another bound I can use $ln(x)leq x-1$. Then I will need to evaluate $int_0^inftyx I_0(x)^2 e^-a x^2 dx$. Do you have an answer for this as well?
– Susan_Math123
Jul 26 at 16:30




@JackD'Aurizio To get another bound I can use $ln(x)leq x-1$. Then I will need to evaluate $int_0^inftyx I_0(x)^2 e^-a x^2 dx$. Do you have an answer for this as well?
– Susan_Math123
Jul 26 at 16:30












Such bound is much worse than the previous one, but in any case $$int_0^+inftyx I_0(x)^2 e^-ax^2,dx = frac12a e^1/(2a) I_0(1/(2a)).$$
– Jack D'Aurizio♦
Jul 26 at 16:42




Such bound is much worse than the previous one, but in any case $$int_0^+inftyx I_0(x)^2 e^-ax^2,dx = frac12a e^1/(2a) I_0(1/(2a)).$$
– Jack D'Aurizio♦
Jul 26 at 16:42















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