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Use De Moivre–Laplace to approximate $1 - sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright)$
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I am trying to use De Moivre–Laplace theorem to approximate
$$1 - sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright)$$
The idea of an approximation is that we don't have the sum term which is difficult to calculate if $n$ is high.
Using the De Moivre–Laplace theorem gets us that: $$n choose k p^k(1-p)^n-k approx frac1sqrt2 pi np(1-p)e^-frac(k-np)^22np(1-p)$$
Now we see that
beginalign
F &= 1 - sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright) \&approx 1 - int_-infty^infty frac1sqrt2 pi np(1-p)e^-frac(x-np)^22np(1-p)log_2left(1+left(fracp1-pright)^n-2xright) dx
endalign
my calculation is inspired by Entropy of a binomial distribution
If one has an other suggestion to approximate $F$ or get a closed for i would like to hear those. So far i've tried approximating $F$ with a least squares method using a tanh function as the fit function.
calculus integration sequences-and-series indefinite-integrals closed-form
asked Aug 2 at 10:06
Kees Til
562617
This question has an open bounty worth +50
reputation from Kees Til ending ending at 2018-08-11 10:21:15Z">in 3 days.
Looking for an answer drawing from credible and/or official sources.
 |Â
show 8 more comments
up vote
3
down vote
favorite
I am trying to use De Moivre–Laplace theorem to approximate
$$1 - sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright)$$
The idea of an approximation is that we don't have the sum term which is difficult to calculate if $n$ is high.
Using the De Moivre–Laplace theorem gets us that: $$n choose k p^k(1-p)^n-k approx frac1sqrt2 pi np(1-p)e^-frac(k-np)^22np(1-p)$$
Now we see that
beginalign
F &= 1 - sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright) \&approx 1 - int_-infty^infty frac1sqrt2 pi np(1-p)e^-frac(x-np)^22np(1-p)log_2left(1+left(fracp1-pright)^n-2xright) dx
endalign
my calculation is inspired by Entropy of a binomial distribution
If one has an other suggestion to approximate $F$ or get a closed for i would like to hear those. So far i've tried approximating $F$ with a least squares method using a tanh function as the fit function.
calculus integration sequences-and-series indefinite-integrals closed-form
asked Aug 2 at 10:06
Kees Til
562617
This question has an open bounty worth +50
reputation from Kees Til ending ending at 2018-08-11 10:21:15Z">in 3 days.
Looking for an answer drawing from credible and/or official sources.
1
Missed $dp$ in the integral
– Yuri Negometyanov
Aug 4 at 20:22
1
I didnt get it. There is something seriously wrong here. The sum formula is eventually independent of $k$ but dependent on $p$ and the final integral is independent of $p$ but dependent on $k$. What does $k$ mean after the integration? sounds strange right?
– Seyhmus Güngören
Aug 4 at 21:06
1
I guess $k$ is eventually $x$ and the integration over $p$ is probably a typo. I do not get the limit $-infty..+infty$ though when originally we have $0..n$.
– Diger
Aug 4 at 21:12
1
Actually it doesnt matter. So for $p<0.5$ use the approximation $log(1+y)approx y$ and for large $y$, we have $log(1+y)approx log(y)$. One more thing you have $y=a^(f(x))$ and you can write this as $exp(a,f(x))$, then in the integral you have the multiplication of two exponentials. I think these forms are well studied.
– Seyhmus Güngören
Aug 4 at 22:07
1
take $p$ small enough and $n$ large enough. I think it should work. Btw. It will be very good, if you could put the results that you already have to the question description. Plus: the answer of wolfram.alpha for the evaluation of this integral, just the link.
– Seyhmus Güngören
Aug 4 at 22:28
 |Â
show 8 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am trying to use De Moivre–Laplace theorem to approximate
$$1 - sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright)$$
The idea of an approximation is that we don't have the sum term which is difficult to calculate if $n$ is high.
Using the De Moivre–Laplace theorem gets us that: $$n choose k p^k(1-p)^n-k approx frac1sqrt2 pi np(1-p)e^-frac(k-np)^22np(1-p)$$
Now we see that
beginalign
F &= 1 - sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright) \&approx 1 - int_-infty^infty frac1sqrt2 pi np(1-p)e^-frac(x-np)^22np(1-p)log_2left(1+left(fracp1-pright)^n-2xright) dx
endalign
my calculation is inspired by Entropy of a binomial distribution
If one has an other suggestion to approximate $F$ or get a closed for i would like to hear those. So far i've tried approximating $F$ with a least squares method using a tanh function as the fit function.
calculus integration sequences-and-series indefinite-integrals closed-form
asked Aug 2 at 10:06
Kees Til
562617
I am trying to use De Moivre–Laplace theorem to approximate
$$1 - sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright)$$
The idea of an approximation is that we don't have the sum term which is difficult to calculate if $n$ is high.
Using the De Moivre–Laplace theorem gets us that: $$n choose k p^k(1-p)^n-k approx frac1sqrt2 pi np(1-p)e^-frac(k-np)^22np(1-p)$$
Now we see that
beginalign
F &= 1 - sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright) \&approx 1 - int_-infty^infty frac1sqrt2 pi np(1-p)e^-frac(x-np)^22np(1-p)log_2left(1+left(fracp1-pright)^n-2xright) dx
endalign
my calculation is inspired by Entropy of a binomial distribution
If one has an other suggestion to approximate $F$ or get a closed for i would like to hear those. So far i've tried approximating $F$ with a least squares method using a tanh function as the fit function.
calculus integration sequences-and-series indefinite-integrals closed-form
asked Aug 2 at 10:06
Kees Til
562617
edited Aug 4 at 21:34
asked Aug 2 at 10:06
Kees Til
562617
asked Aug 2 at 10:06
Kees Til
562617
asked Aug 2 at 10:06
Kees Til
562617
562617
This question has an open bounty worth +50
reputation from Kees Til ending ending at 2018-08-11 10:21:15Z">in 3 days.
Looking for an answer drawing from credible and/or official sources.
This question has an open bounty worth +50
reputation from Kees Til ending ending at 2018-08-11 10:21:15Z">in 3 days.
Looking for an answer drawing from credible and/or official sources.
1
Missed $dp$ in the integral
– Yuri Negometyanov
Aug 4 at 20:22
1
I didnt get it. There is something seriously wrong here. The sum formula is eventually independent of $k$ but dependent on $p$ and the final integral is independent of $p$ but dependent on $k$. What does $k$ mean after the integration? sounds strange right?
– Seyhmus Güngören
Aug 4 at 21:06
1
I guess $k$ is eventually $x$ and the integration over $p$ is probably a typo. I do not get the limit $-infty..+infty$ though when originally we have $0..n$.
– Diger
Aug 4 at 21:12
1
Actually it doesnt matter. So for $p<0.5$ use the approximation $log(1+y)approx y$ and for large $y$, we have $log(1+y)approx log(y)$. One more thing you have $y=a^(f(x))$ and you can write this as $exp(a,f(x))$, then in the integral you have the multiplication of two exponentials. I think these forms are well studied.
– Seyhmus Güngören
Aug 4 at 22:07
1
take $p$ small enough and $n$ large enough. I think it should work. Btw. It will be very good, if you could put the results that you already have to the question description. Plus: the answer of wolfram.alpha for the evaluation of this integral, just the link.
– Seyhmus Güngören
Aug 4 at 22:28
 |Â
show 8 more comments
1
Missed $dp$ in the integral
– Yuri Negometyanov
Aug 4 at 20:22
1
I didnt get it. There is something seriously wrong here. The sum formula is eventually independent of $k$ but dependent on $p$ and the final integral is independent of $p$ but dependent on $k$. What does $k$ mean after the integration? sounds strange right?
– Seyhmus Güngören
Aug 4 at 21:06
1
I guess $k$ is eventually $x$ and the integration over $p$ is probably a typo. I do not get the limit $-infty..+infty$ though when originally we have $0..n$.
– Diger
Aug 4 at 21:12
1
Actually it doesnt matter. So for $p<0.5$ use the approximation $log(1+y)approx y$ and for large $y$, we have $log(1+y)approx log(y)$. One more thing you have $y=a^(f(x))$ and you can write this as $exp(a,f(x))$, then in the integral you have the multiplication of two exponentials. I think these forms are well studied.
– Seyhmus Güngören
Aug 4 at 22:07
1
take $p$ small enough and $n$ large enough. I think it should work. Btw. It will be very good, if you could put the results that you already have to the question description. Plus: the answer of wolfram.alpha for the evaluation of this integral, just the link.
– Seyhmus Güngören
Aug 4 at 22:28
1
1
Missed $dp$ in the integral
– Yuri Negometyanov
Aug 4 at 20:22
Missed $dp$ in the integral
– Yuri Negometyanov
Aug 4 at 20:22
1
1
I didnt get it. There is something seriously wrong here. The sum formula is eventually independent of $k$ but dependent on $p$ and the final integral is independent of $p$ but dependent on $k$. What does $k$ mean after the integration? sounds strange right?
– Seyhmus Güngören
Aug 4 at 21:06
I didnt get it. There is something seriously wrong here. The sum formula is eventually independent of $k$ but dependent on $p$ and the final integral is independent of $p$ but dependent on $k$. What does $k$ mean after the integration? sounds strange right?
– Seyhmus Güngören
Aug 4 at 21:06
1
1
I guess $k$ is eventually $x$ and the integration over $p$ is probably a typo. I do not get the limit $-infty..+infty$ though when originally we have $0..n$.
– Diger
Aug 4 at 21:12
I guess $k$ is eventually $x$ and the integration over $p$ is probably a typo. I do not get the limit $-infty..+infty$ though when originally we have $0..n$.
– Diger
Aug 4 at 21:12
1
1
Actually it doesnt matter. So for $p<0.5$ use the approximation $log(1+y)approx y$ and for large $y$, we have $log(1+y)approx log(y)$. One more thing you have $y=a^(f(x))$ and you can write this as $exp(a,f(x))$, then in the integral you have the multiplication of two exponentials. I think these forms are well studied.
– Seyhmus Güngören
Aug 4 at 22:07
Actually it doesnt matter. So for $p<0.5$ use the approximation $log(1+y)approx y$ and for large $y$, we have $log(1+y)approx log(y)$. One more thing you have $y=a^(f(x))$ and you can write this as $exp(a,f(x))$, then in the integral you have the multiplication of two exponentials. I think these forms are well studied.
– Seyhmus Güngören
Aug 4 at 22:07
1
1
take $p$ small enough and $n$ large enough. I think it should work. Btw. It will be very good, if you could put the results that you already have to the question description. Plus: the answer of wolfram.alpha for the evaluation of this integral, just the link.
– Seyhmus Güngören
Aug 4 at 22:28
take $p$ small enough and $n$ large enough. I think it should work. Btw. It will be very good, if you could put the results that you already have to the question description. Plus: the answer of wolfram.alpha for the evaluation of this integral, just the link.
– Seyhmus Güngören
Aug 4 at 22:28
 |Â
show 8 more comments
1 Answer
1
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up vote
2
down vote
The expression looks very much like the Bernstein approximation of a function ($1-f(x)$) on $[0,1]$. But the argument (in fact the degree $n-2k$) of $log$ function ruins everything.
Here is a quick idea. Denote
$$
y(p)=sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright).
$$
Let us assume that we can represent $y(p)$ in the form $y(p)=sum_m=0^infty y_m p^m$, where $y_m$ are constants not depending on $p$.
Note that $y(p)=y(1-p)$. Let us consider the equation
$$
y(p)+y(1-p)=f(p).
tageq1labeleq1
$$
Although we can write out the expression for $f(p)$, let us think that
we don't know how $f(p)$ looks like. But for sure, $f(p)$ must
satisfy $f(p)=f(1-p)$. It is know (see for example http://eqworld.ipmnet.ru/en/solutions/fe/fe1116.pdf) that equations like eqrefeq1 have a solution, for example,
$$
tageq2labeleq2
y(p)=f(p) sin^2(pi p over 2).
$$
By expanding $sin^2(pi p over 2)$ into the Maclaurin series we get
$$
y(p)=f(p) sum_m=1^infty (-1)^m+1 2^2m-1 over (2m)!
p^2m pi^2m over 2^2m.
$$
Let us assume that $f(p)$ is an analytic function i.e. $f(p)=sum_m=0^infty
f^(m)(0)over m! p^m$. By writing eqrefeq2 in the series form we have:
$$
sum_m=0^infty y_m p^m
=
left (
sum_m=0^infty
f^(m)(0)over m! p^m
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
From this relation it may be possible to find the expressions
for $f^(m)(0)$ through $y_m$ by equating the coefficients at $p^m$. If this works out, we go back to the right part of eqrefeq2 and try to find how many terms in the product
$$
left (
f^(0)(0)
+
f^(1)(0) p
+
f^(2)(0) p over 2
+
dots
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
yield the approximate value.
answered 12 hours ago
rrv
463
so i can't use this because of the "n-2k" term?
– Kees Til
9 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The expression looks very much like the Bernstein approximation of a function ($1-f(x)$) on $[0,1]$. But the argument (in fact the degree $n-2k$) of $log$ function ruins everything.
Here is a quick idea. Denote
$$
y(p)=sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright).
$$
Let us assume that we can represent $y(p)$ in the form $y(p)=sum_m=0^infty y_m p^m$, where $y_m$ are constants not depending on $p$.
Note that $y(p)=y(1-p)$. Let us consider the equation
$$
y(p)+y(1-p)=f(p).
tageq1labeleq1
$$
Although we can write out the expression for $f(p)$, let us think that
we don't know how $f(p)$ looks like. But for sure, $f(p)$ must
satisfy $f(p)=f(1-p)$. It is know (see for example http://eqworld.ipmnet.ru/en/solutions/fe/fe1116.pdf) that equations like eqrefeq1 have a solution, for example,
$$
tageq2labeleq2
y(p)=f(p) sin^2(pi p over 2).
$$
By expanding $sin^2(pi p over 2)$ into the Maclaurin series we get
$$
y(p)=f(p) sum_m=1^infty (-1)^m+1 2^2m-1 over (2m)!
p^2m pi^2m over 2^2m.
$$
Let us assume that $f(p)$ is an analytic function i.e. $f(p)=sum_m=0^infty
f^(m)(0)over m! p^m$. By writing eqrefeq2 in the series form we have:
$$
sum_m=0^infty y_m p^m
=
left (
sum_m=0^infty
f^(m)(0)over m! p^m
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
From this relation it may be possible to find the expressions
for $f^(m)(0)$ through $y_m$ by equating the coefficients at $p^m$. If this works out, we go back to the right part of eqrefeq2 and try to find how many terms in the product
$$
left (
f^(0)(0)
+
f^(1)(0) p
+
f^(2)(0) p over 2
+
dots
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
yield the approximate value.
answered 12 hours ago
rrv
463
so i can't use this because of the "n-2k" term?
– Kees Til
9 hours ago
add a comment |Â
up vote
2
down vote
The expression looks very much like the Bernstein approximation of a function ($1-f(x)$) on $[0,1]$. But the argument (in fact the degree $n-2k$) of $log$ function ruins everything.
Here is a quick idea. Denote
$$
y(p)=sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright).
$$
Let us assume that we can represent $y(p)$ in the form $y(p)=sum_m=0^infty y_m p^m$, where $y_m$ are constants not depending on $p$.
Note that $y(p)=y(1-p)$. Let us consider the equation
$$
y(p)+y(1-p)=f(p).
tageq1labeleq1
$$
Although we can write out the expression for $f(p)$, let us think that
we don't know how $f(p)$ looks like. But for sure, $f(p)$ must
satisfy $f(p)=f(1-p)$. It is know (see for example http://eqworld.ipmnet.ru/en/solutions/fe/fe1116.pdf) that equations like eqrefeq1 have a solution, for example,
$$
tageq2labeleq2
y(p)=f(p) sin^2(pi p over 2).
$$
By expanding $sin^2(pi p over 2)$ into the Maclaurin series we get
$$
y(p)=f(p) sum_m=1^infty (-1)^m+1 2^2m-1 over (2m)!
p^2m pi^2m over 2^2m.
$$
Let us assume that $f(p)$ is an analytic function i.e. $f(p)=sum_m=0^infty
f^(m)(0)over m! p^m$. By writing eqrefeq2 in the series form we have:
$$
sum_m=0^infty y_m p^m
=
left (
sum_m=0^infty
f^(m)(0)over m! p^m
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
From this relation it may be possible to find the expressions
for $f^(m)(0)$ through $y_m$ by equating the coefficients at $p^m$. If this works out, we go back to the right part of eqrefeq2 and try to find how many terms in the product
$$
left (
f^(0)(0)
+
f^(1)(0) p
+
f^(2)(0) p over 2
+
dots
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
yield the approximate value.
answered 12 hours ago
rrv
463
so i can't use this because of the "n-2k" term?
– Kees Til
9 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The expression looks very much like the Bernstein approximation of a function ($1-f(x)$) on $[0,1]$. But the argument (in fact the degree $n-2k$) of $log$ function ruins everything.
Here is a quick idea. Denote
$$
y(p)=sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright).
$$
Let us assume that we can represent $y(p)$ in the form $y(p)=sum_m=0^infty y_m p^m$, where $y_m$ are constants not depending on $p$.
Note that $y(p)=y(1-p)$. Let us consider the equation
$$
y(p)+y(1-p)=f(p).
tageq1labeleq1
$$
Although we can write out the expression for $f(p)$, let us think that
we don't know how $f(p)$ looks like. But for sure, $f(p)$ must
satisfy $f(p)=f(1-p)$. It is know (see for example http://eqworld.ipmnet.ru/en/solutions/fe/fe1116.pdf) that equations like eqrefeq1 have a solution, for example,
$$
tageq2labeleq2
y(p)=f(p) sin^2(pi p over 2).
$$
By expanding $sin^2(pi p over 2)$ into the Maclaurin series we get
$$
y(p)=f(p) sum_m=1^infty (-1)^m+1 2^2m-1 over (2m)!
p^2m pi^2m over 2^2m.
$$
Let us assume that $f(p)$ is an analytic function i.e. $f(p)=sum_m=0^infty
f^(m)(0)over m! p^m$. By writing eqrefeq2 in the series form we have:
$$
sum_m=0^infty y_m p^m
=
left (
sum_m=0^infty
f^(m)(0)over m! p^m
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
From this relation it may be possible to find the expressions
for $f^(m)(0)$ through $y_m$ by equating the coefficients at $p^m$. If this works out, we go back to the right part of eqrefeq2 and try to find how many terms in the product
$$
left (
f^(0)(0)
+
f^(1)(0) p
+
f^(2)(0) p over 2
+
dots
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
yield the approximate value.
answered 12 hours ago
rrv
463
The expression looks very much like the Bernstein approximation of a function ($1-f(x)$) on $[0,1]$. But the argument (in fact the degree $n-2k$) of $log$ function ruins everything.
Here is a quick idea. Denote
$$
y(p)=sum_k=0^n n choose k p^k(1-p)^n-k logleft(1+left(fracp1-pright)^n-2kright).
$$
Let us assume that we can represent $y(p)$ in the form $y(p)=sum_m=0^infty y_m p^m$, where $y_m$ are constants not depending on $p$.
Note that $y(p)=y(1-p)$. Let us consider the equation
$$
y(p)+y(1-p)=f(p).
tageq1labeleq1
$$
Although we can write out the expression for $f(p)$, let us think that
we don't know how $f(p)$ looks like. But for sure, $f(p)$ must
satisfy $f(p)=f(1-p)$. It is know (see for example http://eqworld.ipmnet.ru/en/solutions/fe/fe1116.pdf) that equations like eqrefeq1 have a solution, for example,
$$
tageq2labeleq2
y(p)=f(p) sin^2(pi p over 2).
$$
By expanding $sin^2(pi p over 2)$ into the Maclaurin series we get
$$
y(p)=f(p) sum_m=1^infty (-1)^m+1 2^2m-1 over (2m)!
p^2m pi^2m over 2^2m.
$$
Let us assume that $f(p)$ is an analytic function i.e. $f(p)=sum_m=0^infty
f^(m)(0)over m! p^m$. By writing eqrefeq2 in the series form we have:
$$
sum_m=0^infty y_m p^m
=
left (
sum_m=0^infty
f^(m)(0)over m! p^m
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
From this relation it may be possible to find the expressions
for $f^(m)(0)$ through $y_m$ by equating the coefficients at $p^m$. If this works out, we go back to the right part of eqrefeq2 and try to find how many terms in the product
$$
left (
f^(0)(0)
+
f^(1)(0) p
+
f^(2)(0) p over 2
+
dots
right )
left (
sum_m=1^infty (-1)^m+1 over (2m)!
p^2m pi^2m over 2
right ).
$$
yield the approximate value.
answered 12 hours ago
rrv
463
answered 12 hours ago
rrv
463
answered 12 hours ago
rrv
463
answered 12 hours ago
rrv
463
463
so i can't use this because of the "n-2k" term?
– Kees Til
9 hours ago
add a comment |Â
so i can't use this because of the "n-2k" term?
– Kees Til
9 hours ago
so i can't use this because of the "n-2k" term?
– Kees Til
9 hours ago
so i can't use this because of the "n-2k" term?
– Kees Til
9 hours ago
add a comment |Â
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1
Missed $dp$ in the integral
– Yuri Negometyanov
Aug 4 at 20:22
1
I didnt get it. There is something seriously wrong here. The sum formula is eventually independent of $k$ but dependent on $p$ and the final integral is independent of $p$ but dependent on $k$. What does $k$ mean after the integration? sounds strange right?
– Seyhmus Güngören
Aug 4 at 21:06
1
I guess $k$ is eventually $x$ and the integration over $p$ is probably a typo. I do not get the limit $-infty..+infty$ though when originally we have $0..n$.
– Diger
Aug 4 at 21:12
1
Actually it doesnt matter. So for $p<0.5$ use the approximation $log(1+y)approx y$ and for large $y$, we have $log(1+y)approx log(y)$. One more thing you have $y=a^(f(x))$ and you can write this as $exp(a,f(x))$, then in the integral you have the multiplication of two exponentials. I think these forms are well studied.
– Seyhmus Güngören
Aug 4 at 22:07
1
take $p$ small enough and $n$ large enough. I think it should work. Btw. It will be very good, if you could put the results that you already have to the question description. Plus: the answer of wolfram.alpha for the evaluation of this integral, just the link.
– Seyhmus Güngören
Aug 4 at 22:28