Mixing
Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $
$begingroup$
I'm trying to calculate this integral:
$$ int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $$
For $a > 0$.
This is what I did:
$$ I(a) = int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx \
I'(a) = int_{0}^{pi/2} frac{2asin^2{x}}{(a^2sin^2{x} + cos^2{x})} dx \
I'(a) = int_{0}^{pi/2} frac{2a}{(a^2 + frac{cos^2{x}}{sin^2{x}})} dx \
I'(a) = 2a int_{0}^{pi/2} frac{1}{a^2 + cot^2{x}}dx \
I'(a) = frac{2}{a} int_{0}^{pi/2} frac{1}{1 + frac{cot^2{x}}{a^2}}dx $$
Here I tried to substitude $frac{cot^2{x}}{a^2} = t$. This should lead me to $arctan(something)$ (I write $arctan{x} = tan^{-1}{x}$.)
But I got stuck.
After I get the derivative I should integrate it back to get $I(a)$. Also there are steps that need some prepositions to be checked. Could one help me with this integral and also to clarify the steps that need special attention; such as the second step, where I can get the derivative $I'(a)$ only if the function under the integral can be differentiated? It must be possible to find its derivative.
real-analysis calculus analysis
edited Jan 17 at 16:01
KM101
6,0251525
asked Jan 17 at 15:37
CoupeauCoupeau
1296
$endgroup$
add a comment |
$begingroup$
I'm trying to calculate this integral:
$$ int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $$
For $a > 0$.
This is what I did:
$$ I(a) = int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx \
I'(a) = int_{0}^{pi/2} frac{2asin^2{x}}{(a^2sin^2{x} + cos^2{x})} dx \
I'(a) = int_{0}^{pi/2} frac{2a}{(a^2 + frac{cos^2{x}}{sin^2{x}})} dx \
I'(a) = 2a int_{0}^{pi/2} frac{1}{a^2 + cot^2{x}}dx \
I'(a) = frac{2}{a} int_{0}^{pi/2} frac{1}{1 + frac{cot^2{x}}{a^2}}dx $$
Here I tried to substitude $frac{cot^2{x}}{a^2} = t$. This should lead me to $arctan(something)$ (I write $arctan{x} = tan^{-1}{x}$.)
But I got stuck.
After I get the derivative I should integrate it back to get $I(a)$. Also there are steps that need some prepositions to be checked. Could one help me with this integral and also to clarify the steps that need special attention; such as the second step, where I can get the derivative $I'(a)$ only if the function under the integral can be differentiated? It must be possible to find its derivative.
real-analysis calculus analysis
edited Jan 17 at 16:01
KM101
6,0251525
asked Jan 17 at 15:37
CoupeauCoupeau
1296
$endgroup$
add a comment |
$begingroup$
I'm trying to calculate this integral:
$$ int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $$
For $a > 0$.
This is what I did:
$$ I(a) = int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx \
I'(a) = int_{0}^{pi/2} frac{2asin^2{x}}{(a^2sin^2{x} + cos^2{x})} dx \
I'(a) = int_{0}^{pi/2} frac{2a}{(a^2 + frac{cos^2{x}}{sin^2{x}})} dx \
I'(a) = 2a int_{0}^{pi/2} frac{1}{a^2 + cot^2{x}}dx \
I'(a) = frac{2}{a} int_{0}^{pi/2} frac{1}{1 + frac{cot^2{x}}{a^2}}dx $$
Here I tried to substitude $frac{cot^2{x}}{a^2} = t$. This should lead me to $arctan(something)$ (I write $arctan{x} = tan^{-1}{x}$.)
But I got stuck.
After I get the derivative I should integrate it back to get $I(a)$. Also there are steps that need some prepositions to be checked. Could one help me with this integral and also to clarify the steps that need special attention; such as the second step, where I can get the derivative $I'(a)$ only if the function under the integral can be differentiated? It must be possible to find its derivative.
real-analysis calculus analysis
edited Jan 17 at 16:01
KM101
6,0251525
asked Jan 17 at 15:37
CoupeauCoupeau
1296
$endgroup$
I'm trying to calculate this integral:
$$ int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $$
For $a > 0$.
This is what I did:
$$ I(a) = int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx \
I'(a) = int_{0}^{pi/2} frac{2asin^2{x}}{(a^2sin^2{x} + cos^2{x})} dx \
I'(a) = int_{0}^{pi/2} frac{2a}{(a^2 + frac{cos^2{x}}{sin^2{x}})} dx \
I'(a) = 2a int_{0}^{pi/2} frac{1}{a^2 + cot^2{x}}dx \
I'(a) = frac{2}{a} int_{0}^{pi/2} frac{1}{1 + frac{cot^2{x}}{a^2}}dx $$
Here I tried to substitude $frac{cot^2{x}}{a^2} = t$. This should lead me to $arctan(something)$ (I write $arctan{x} = tan^{-1}{x}$.)
But I got stuck.
After I get the derivative I should integrate it back to get $I(a)$. Also there are steps that need some prepositions to be checked. Could one help me with this integral and also to clarify the steps that need special attention; such as the second step, where I can get the derivative $I'(a)$ only if the function under the integral can be differentiated? It must be possible to find its derivative.
real-analysis calculus analysis
real-analysis calculus analysis
edited Jan 17 at 16:01
KM101
6,0251525
asked Jan 17 at 15:37
CoupeauCoupeau
1296
edited Jan 17 at 16:01
KM101
6,0251525
asked Jan 17 at 15:37
CoupeauCoupeau
1296
edited Jan 17 at 16:01
KM101
6,0251525
edited Jan 17 at 16:01
KM101
6,0251525
edited Jan 17 at 16:01
KM101
6,0251525
6,0251525
asked Jan 17 at 15:37
CoupeauCoupeau
1296
asked Jan 17 at 15:37
CoupeauCoupeau
1296
asked Jan 17 at 15:37
CoupeauCoupeau
1296
1296
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that, under $cot xto x$,
begin{eqnarray*}
I'(a)& = &2a int_{0}^{pi/2} frac{1}{a^2 + cot^2{x}}dx\
&=&2aint_0^inftyfrac{1}{(a^2+x^2)(1+x^2)}dx\
&=&frac{2a}{1-a^2}int_0^inftybigg(frac{1}{a^2+x^2}-frac{1}{1+x^2}bigg)dx
&=&frac{pi}{1+a}.
end{eqnarray*}
which can be easily handled. So
$$ I(1)-I(a)=int_a^1frac{pi}{1+t}dt=pi(ln2-ln(1+a))$$
and hence
$$ I(a)=-piln2+piln(1+a)=pilnfrac{1+a}{2}. $$
since $I(1)=0$.
answered Jan 17 at 16:09
xpaulxpaul
22.9k24455
$endgroup$
$begingroup$
Thank you.Only, I do not understand how you got $I(0) = 2 int_{0}^{frac{pi}{2}} ln{cos{x}} dx$ to be $-pi ln2$. I did get the same resoult after calculating $I(a) = pi ln{(1+a)} + C$ and $I(1) = 0$. Could you explain a little further how you calculated this integal?
$endgroup$
– Coupeau
Jan 18 at 20:00
$begingroup$
Yes, it will be better if using $I(1)=0$.
$endgroup$
– xpaul
Jan 18 at 20:23
add a comment |
$begingroup$
you can also use the series of $ln x$ to find it
$$begin{aligned}
int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x
& = int_{0}^{pi/2} ln(1+(a^2-1)sin^2{x}) mathrm{d}x\
& = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} int_{0}^{pi/2}(a^2-1)^nsin^{2n}{x} mathrm{d}x\
& = pi sum_{n=1}^{infty} frac{(-1)^{n-1}}{2n} frac{(2n-1)!!}{(2n)!!} (a^2-1)^n
end{aligned}$$
where you need wallis' integral here, then using this identity
$$operatorname{arsinh} x = ln(2x) + sumlimits_{n = 1}^infty {left( { - 1} right)^{n - 1} frac{{left( {2n - 1} right)!!}}{{2nleft( {2n} right)!!}}} frac{1}{{x^{2n} }}$$
let $x=1/sqrt{a^2-1}$ and you will find the answer
$$int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x = pileft( operatorname{arsinh} frac1{sqrt{a^2-1}} - lnfrac2{sqrt{a^2-1}} right) = pilnfrac{a+1}{2}$$
answered Jan 17 at 16:31
NanayajitzukiNanayajitzuki
3185
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that, under $cot xto x$,
begin{eqnarray*}
I'(a)& = &2a int_{0}^{pi/2} frac{1}{a^2 + cot^2{x}}dx\
&=&2aint_0^inftyfrac{1}{(a^2+x^2)(1+x^2)}dx\
&=&frac{2a}{1-a^2}int_0^inftybigg(frac{1}{a^2+x^2}-frac{1}{1+x^2}bigg)dx
&=&frac{pi}{1+a}.
end{eqnarray*}
which can be easily handled. So
$$ I(1)-I(a)=int_a^1frac{pi}{1+t}dt=pi(ln2-ln(1+a))$$
and hence
$$ I(a)=-piln2+piln(1+a)=pilnfrac{1+a}{2}. $$
since $I(1)=0$.
answered Jan 17 at 16:09
xpaulxpaul
22.9k24455
$endgroup$
$begingroup$
Thank you.Only, I do not understand how you got $I(0) = 2 int_{0}^{frac{pi}{2}} ln{cos{x}} dx$ to be $-pi ln2$. I did get the same resoult after calculating $I(a) = pi ln{(1+a)} + C$ and $I(1) = 0$. Could you explain a little further how you calculated this integal?
$endgroup$
– Coupeau
Jan 18 at 20:00
$begingroup$
Yes, it will be better if using $I(1)=0$.
$endgroup$
– xpaul
Jan 18 at 20:23
add a comment |
$begingroup$
Note that, under $cot xto x$,
begin{eqnarray*}
I'(a)& = &2a int_{0}^{pi/2} frac{1}{a^2 + cot^2{x}}dx\
&=&2aint_0^inftyfrac{1}{(a^2+x^2)(1+x^2)}dx\
&=&frac{2a}{1-a^2}int_0^inftybigg(frac{1}{a^2+x^2}-frac{1}{1+x^2}bigg)dx
&=&frac{pi}{1+a}.
end{eqnarray*}
which can be easily handled. So
$$ I(1)-I(a)=int_a^1frac{pi}{1+t}dt=pi(ln2-ln(1+a))$$
and hence
$$ I(a)=-piln2+piln(1+a)=pilnfrac{1+a}{2}. $$
since $I(1)=0$.
answered Jan 17 at 16:09
xpaulxpaul
22.9k24455
$endgroup$
$begingroup$
Thank you.Only, I do not understand how you got $I(0) = 2 int_{0}^{frac{pi}{2}} ln{cos{x}} dx$ to be $-pi ln2$. I did get the same resoult after calculating $I(a) = pi ln{(1+a)} + C$ and $I(1) = 0$. Could you explain a little further how you calculated this integal?
$endgroup$
– Coupeau
Jan 18 at 20:00
$begingroup$
Yes, it will be better if using $I(1)=0$.
$endgroup$
– xpaul
Jan 18 at 20:23
add a comment |
$begingroup$
Note that, under $cot xto x$,
begin{eqnarray*}
I'(a)& = &2a int_{0}^{pi/2} frac{1}{a^2 + cot^2{x}}dx\
&=&2aint_0^inftyfrac{1}{(a^2+x^2)(1+x^2)}dx\
&=&frac{2a}{1-a^2}int_0^inftybigg(frac{1}{a^2+x^2}-frac{1}{1+x^2}bigg)dx
&=&frac{pi}{1+a}.
end{eqnarray*}
which can be easily handled. So
$$ I(1)-I(a)=int_a^1frac{pi}{1+t}dt=pi(ln2-ln(1+a))$$
and hence
$$ I(a)=-piln2+piln(1+a)=pilnfrac{1+a}{2}. $$
since $I(1)=0$.
answered Jan 17 at 16:09
xpaulxpaul
22.9k24455
$endgroup$
Note that, under $cot xto x$,
begin{eqnarray*}
I'(a)& = &2a int_{0}^{pi/2} frac{1}{a^2 + cot^2{x}}dx\
&=&2aint_0^inftyfrac{1}{(a^2+x^2)(1+x^2)}dx\
&=&frac{2a}{1-a^2}int_0^inftybigg(frac{1}{a^2+x^2}-frac{1}{1+x^2}bigg)dx
&=&frac{pi}{1+a}.
end{eqnarray*}
which can be easily handled. So
$$ I(1)-I(a)=int_a^1frac{pi}{1+t}dt=pi(ln2-ln(1+a))$$
and hence
$$ I(a)=-piln2+piln(1+a)=pilnfrac{1+a}{2}. $$
since $I(1)=0$.
answered Jan 17 at 16:09
xpaulxpaul
22.9k24455
edited Jan 18 at 20:25
answered Jan 17 at 16:09
xpaulxpaul
22.9k24455
answered Jan 17 at 16:09
xpaulxpaul
22.9k24455
answered Jan 17 at 16:09
xpaulxpaul
22.9k24455
22.9k24455
$begingroup$
Thank you.Only, I do not understand how you got $I(0) = 2 int_{0}^{frac{pi}{2}} ln{cos{x}} dx$ to be $-pi ln2$. I did get the same resoult after calculating $I(a) = pi ln{(1+a)} + C$ and $I(1) = 0$. Could you explain a little further how you calculated this integal?
$endgroup$
– Coupeau
Jan 18 at 20:00
$begingroup$
Yes, it will be better if using $I(1)=0$.
$endgroup$
– xpaul
Jan 18 at 20:23
add a comment |
$begingroup$
Thank you.Only, I do not understand how you got $I(0) = 2 int_{0}^{frac{pi}{2}} ln{cos{x}} dx$ to be $-pi ln2$. I did get the same resoult after calculating $I(a) = pi ln{(1+a)} + C$ and $I(1) = 0$. Could you explain a little further how you calculated this integal?
$endgroup$
– Coupeau
Jan 18 at 20:00
$begingroup$
Yes, it will be better if using $I(1)=0$.
$endgroup$
– xpaul
Jan 18 at 20:23
$begingroup$
Thank you.Only, I do not understand how you got $I(0) = 2 int_{0}^{frac{pi}{2}} ln{cos{x}} dx$ to be $-pi ln2$. I did get the same resoult after calculating $I(a) = pi ln{(1+a)} + C$ and $I(1) = 0$. Could you explain a little further how you calculated this integal?
$endgroup$
– Coupeau
Jan 18 at 20:00
$begingroup$
Thank you.Only, I do not understand how you got $I(0) = 2 int_{0}^{frac{pi}{2}} ln{cos{x}} dx$ to be $-pi ln2$. I did get the same resoult after calculating $I(a) = pi ln{(1+a)} + C$ and $I(1) = 0$. Could you explain a little further how you calculated this integal?
$endgroup$
– Coupeau
Jan 18 at 20:00
$begingroup$
Yes, it will be better if using $I(1)=0$.
$endgroup$
– xpaul
Jan 18 at 20:23
$begingroup$
Yes, it will be better if using $I(1)=0$.
$endgroup$
– xpaul
Jan 18 at 20:23
add a comment |
$begingroup$
you can also use the series of $ln x$ to find it
$$begin{aligned}
int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x
& = int_{0}^{pi/2} ln(1+(a^2-1)sin^2{x}) mathrm{d}x\
& = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} int_{0}^{pi/2}(a^2-1)^nsin^{2n}{x} mathrm{d}x\
& = pi sum_{n=1}^{infty} frac{(-1)^{n-1}}{2n} frac{(2n-1)!!}{(2n)!!} (a^2-1)^n
end{aligned}$$
where you need wallis' integral here, then using this identity
$$operatorname{arsinh} x = ln(2x) + sumlimits_{n = 1}^infty {left( { - 1} right)^{n - 1} frac{{left( {2n - 1} right)!!}}{{2nleft( {2n} right)!!}}} frac{1}{{x^{2n} }}$$
let $x=1/sqrt{a^2-1}$ and you will find the answer
$$int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x = pileft( operatorname{arsinh} frac1{sqrt{a^2-1}} - lnfrac2{sqrt{a^2-1}} right) = pilnfrac{a+1}{2}$$
answered Jan 17 at 16:31
NanayajitzukiNanayajitzuki
3185
$endgroup$
add a comment |
$begingroup$
you can also use the series of $ln x$ to find it
$$begin{aligned}
int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x
& = int_{0}^{pi/2} ln(1+(a^2-1)sin^2{x}) mathrm{d}x\
& = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} int_{0}^{pi/2}(a^2-1)^nsin^{2n}{x} mathrm{d}x\
& = pi sum_{n=1}^{infty} frac{(-1)^{n-1}}{2n} frac{(2n-1)!!}{(2n)!!} (a^2-1)^n
end{aligned}$$
where you need wallis' integral here, then using this identity
$$operatorname{arsinh} x = ln(2x) + sumlimits_{n = 1}^infty {left( { - 1} right)^{n - 1} frac{{left( {2n - 1} right)!!}}{{2nleft( {2n} right)!!}}} frac{1}{{x^{2n} }}$$
let $x=1/sqrt{a^2-1}$ and you will find the answer
$$int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x = pileft( operatorname{arsinh} frac1{sqrt{a^2-1}} - lnfrac2{sqrt{a^2-1}} right) = pilnfrac{a+1}{2}$$
answered Jan 17 at 16:31
NanayajitzukiNanayajitzuki
3185
$endgroup$
add a comment |
$begingroup$
you can also use the series of $ln x$ to find it
$$begin{aligned}
int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x
& = int_{0}^{pi/2} ln(1+(a^2-1)sin^2{x}) mathrm{d}x\
& = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} int_{0}^{pi/2}(a^2-1)^nsin^{2n}{x} mathrm{d}x\
& = pi sum_{n=1}^{infty} frac{(-1)^{n-1}}{2n} frac{(2n-1)!!}{(2n)!!} (a^2-1)^n
end{aligned}$$
where you need wallis' integral here, then using this identity
$$operatorname{arsinh} x = ln(2x) + sumlimits_{n = 1}^infty {left( { - 1} right)^{n - 1} frac{{left( {2n - 1} right)!!}}{{2nleft( {2n} right)!!}}} frac{1}{{x^{2n} }}$$
let $x=1/sqrt{a^2-1}$ and you will find the answer
$$int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x = pileft( operatorname{arsinh} frac1{sqrt{a^2-1}} - lnfrac2{sqrt{a^2-1}} right) = pilnfrac{a+1}{2}$$
answered Jan 17 at 16:31
NanayajitzukiNanayajitzuki
3185
$endgroup$
you can also use the series of $ln x$ to find it
$$begin{aligned}
int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x
& = int_{0}^{pi/2} ln(1+(a^2-1)sin^2{x}) mathrm{d}x\
& = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} int_{0}^{pi/2}(a^2-1)^nsin^{2n}{x} mathrm{d}x\
& = pi sum_{n=1}^{infty} frac{(-1)^{n-1}}{2n} frac{(2n-1)!!}{(2n)!!} (a^2-1)^n
end{aligned}$$
where you need wallis' integral here, then using this identity
$$operatorname{arsinh} x = ln(2x) + sumlimits_{n = 1}^infty {left( { - 1} right)^{n - 1} frac{{left( {2n - 1} right)!!}}{{2nleft( {2n} right)!!}}} frac{1}{{x^{2n} }}$$
let $x=1/sqrt{a^2-1}$ and you will find the answer
$$int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x}) mathrm{d}x = pileft( operatorname{arsinh} frac1{sqrt{a^2-1}} - lnfrac2{sqrt{a^2-1}} right) = pilnfrac{a+1}{2}$$
answered Jan 17 at 16:31
NanayajitzukiNanayajitzuki
3185
edited Jan 17 at 16:37
answered Jan 17 at 16:31
NanayajitzukiNanayajitzuki
3185
answered Jan 17 at 16:31
NanayajitzukiNanayajitzuki
3185
answered Jan 17 at 16:31
NanayajitzukiNanayajitzuki
3185
3185
add a comment |
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
