Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to solve the following integral.

$I(\theta) = \int_0^{\pi}\ln(1+\theta \cos x)dx$ where $|\theta|<1$

share|cite|improve this question

Differentiating under the integral sign yields: $$ I'(\theta) = \int^{\pi}_0 \frac{\cos x}{1+ \theta \cos x} dx .$$

So $$\theta I'(\theta) = \int^{\pi}_0 1 - \frac{1}{1+\theta \cos x} dx= \pi - \int^{\pi}_0 \frac{1}{1+\theta \cos x} dx$$

To deal with the last integral, consider $$ \[\begin{aligned} I & = \int_{0}^{\pi} \frac{1}{a+b\cos{x}}\;{dx} \\& = \int_{0}^{\pi} \frac{1}{a\left(\sin^2\frac{1}{2}x+\cos^2\frac{1}{2}x\right)+b\left(\cos^2\frac{1}{2}x-\sin^2\frac{1}{2}x\right)}\;{dx} \\& =\int_{0}^{\pi}\frac{1}{(a-b)\sin^2\frac{1}{2}x+(a+b)\cos^2\frac{1}{2}x}\;{dx} \\& =\int_{0}^{\pi}\frac{\sec^2{\frac{1}{2}x}}{(a+b)+(a-b)\tan^2\frac{1}{2}x}\;{dx} \\& = 2\int_{0}^{\infty}\frac{1}{(a+b)+(a-b)t^2}\;{dt} \\& = 2\int_{0}^{\infty}\frac{1}{(\sqrt{a+b})^2+(\sqrt{a-b})^2t^2}\;{dt}\\& = \frac{2}{{\sqrt{a^2-b^2}}}\tan^{-1}\bigg(\frac{\sqrt{a-b}}{\sqrt{a+b}}~t \bigg)\bigg|_{0}^{\infty} \\& = \frac{\pi}{\sqrt{a^2-b^2}}.\end{aligned}\] $$

Thus, $$ \theta I'(\theta) = \pi \left(1 - \frac{1}{\sqrt{1-\theta^2}} \right).$$

You can now do some integration of your own to find $I'(\theta).$

share|cite|improve this answer

Note that the integrand always changes sign at $x=\frac\pi2$ for $\theta\ne0$. In fact, this is an even, nonpositive function, since $\cos(\pi-x)=-\cos x$ and since, for $r=\theta\cos x$, $|r|<1$ and $\ln(1+r)+$ $\ln(1-r)=$ $\ln(1-r^2)<0$ $\implies$ $$ \eqalign{ \int_0^\pi~\ln\big(1+\theta\,\cos\,x\big)\;dx &= \int_0^\frac\pi2~\ln\big(1+\theta\cos x\big)\;dx + \int_\frac\pi2^\pi~\ln\big(1+\theta\cos x\big)\;dx \\ &= \int_0^\frac\pi2~\ln\big(1-\theta^2\cos^2 x\big)\;dx \le 0 \,, } $$ with equality iff $\theta=0$. Out of perverse curiosity, let us define, slightly more generally (substituting $a$ for 1, $b$ for $\theta$ and $\theta$ for $x$) $$ I(a,b)=\int_0^\pi\ln\big(a+b\cos\theta\big)\;d\theta. $$ Then $$ \frac{\partial I}{\partial b} =\int_0^\pi \frac{\cos\theta\,d\theta}{a+b\cos\theta} =\frac{\pi}{b}-\frac{a}b\int_0^\pi\frac{d\theta}{a+b\cos\theta} \qquad \text{since} \qquad \frac{b\cos\theta}{a+b\cos\theta} =1-\frac{a }{a+b\cos\theta} . $$ But (thanks to Ragib Zaman): $$ \eqalign{\frac\pi{a} - \frac{b}{a} \, \frac{\partial I}{\partial b} & = \int_0^\pi \frac{d\theta}{a+b\cos{\theta}} \\& = \int_0^\pi \frac{d\theta} {a\left(\sin^2\frac\theta2+\cos^2\frac\theta2\right) +b\left(\cos^2\frac\theta2-\sin^2\frac\theta2\right)} \\& = \int_0^\pi \frac{d\theta} {(a-b)\sin^2\frac\theta2+(a+b)\cos^2\frac\theta2} \\& = \int_0^\pi \frac{\sec^2\frac\theta2\;d\theta} {(a+b)+(a-b)\tan^2\frac\theta2} \\& = 2\int_0^\infty \frac{dt}{(a+b)+(a-b)t^2} \\& = 2\int_0^\infty \frac{dt}{(\sqrt{a+b})^2+(\sqrt{a-b})^2t^2} \\& = \frac{2}{{\sqrt{a^2-b^2}}}~ \left.\tan^{-1} \left( \frac{\sqrt{a-b}}{\sqrt{a+b}}~t \right) \right|_{0}^{\infty} \\& = \frac{\pi}{\sqrt{a^2-b^2}} } $$ so that $$ \frac{\partial I}{\partial b} = \frac\pi{b} \left( 1-\frac{a}{\sqrt{a^2-b^2}} \right) $$ or $$ I(a,b) = \pi \int \frac{db}{b} - \pi a \int \frac{db}{b \sqrt{a^2-b^2}} \,. $$ For $a>|b|>0$, we can use the substitution $b=a\sin\phi,~db=a\cos\phi\,d\phi$ to continue thus: $$ \eqalign{ I(a,b)&=\pi\ln|b| - \pi a \int \frac{d\phi}{\sin\phi} \\ &=\pi\ln|b| - \pi a \int \csc\phi ~d\phi \\ &=\pi\ln|b| + \pi a ~\ln\, \big| \csc\phi + \cot\phi \big| + c \\ &=\pi\ln|b| + \pi a ~\ln\, \left| \frac{1+\cos\phi}{\sin\phi} \right| + c \\ &=\pi\ln|b| + \pi a ~\ln\, \left| \frac{a}{b} +\sqrt{\left( \frac{a}{b} \right)^2-1} \right| + c \,. } $$ Using $I(a,0)=\pi\,\ln a$, we find that $c$ depends on a limit which exists and is $\ln2$ iff $a=1$, $$ \frac{\pi\ln a-c}{a} =\lim_{b\rightarrow0}\,\ln \left| \frac{ \frac{a}{b} +\sqrt{\left( \frac{a}{b} \right)^2-1} }{|b|^{-1/a}} \right| =\lim_{b\rightarrow0}\,\ln \left| b^{1/a-1} \left( a+\sqrt{a^2-b^2} \right) \right| $$ in which case $c=-\pi\ln2$. So for our original problem, $$ \int_0^\pi~\ln\big(1+\theta\,\cos\,x\big)\;dx=I(1,\theta) = \pi \ln\frac{1+\sqrt{1-\theta^2}}{2} \,. $$ As already noted, this exists and is nonpositive for $|\theta|<1$. On the other hand we see from the RHS above that the integral is bounded below by $-\pi\ln2\approx-2.17758609030360$. Here is a plot of the solution using sage, with the factor of $\pi$ removed (in blue), and a comparable function (in red) from an earlier erroneous draft:

#assume(t != 0)
G = plot(log( (1 + sqrt(1-t^2)) / 2 ), (t, -1, 1), color='blue')
G+= text('log((1 + sqrt(1-t^2)) / 2)', (-.6,-.65), color='blue')
G+= plot(      1 - arcsin(t)/t       , (t, -1, 1), color='red')
G+= text(     '1 - arcsin(t)/t'      , (.85,-.06), color='red')

enter image description here

In particular, the graph has a minimum of $-\ln2\approx-0.693147180559945$ at $\theta=\pm1$.

To check the endpoint, I computed an approximate numerical integral, which agrees with the above (the tuple gives the integral and error estimates):

numerical_integral(2*log(sin(x)), 0, pi/2)

$\left(-2.17758608788, 1.09713268507 \times 10^{-06}\right)$

share|cite|improve this answer
$$ I(a,b) = \pi \ln|b| - \pi\frac{a}{b} \int\frac{db}{\sqrt{a^2-b^2}} $$ this equation should be $$ I(a,b) = \pi \ln|b| - \pi a \int\frac{db}{b\sqrt{a^2-b^2}} $$ ???? – Mathlover Feb 29 '12 at 11:41
Thanks...of course! – bgins Feb 29 '12 at 13:47

Did you try that way?

$I(\theta) = \int_0^{\pi}\ln(1+\theta \cos x)dx$

$I'(\theta) = \int_0^{\pi}\frac{\cos x}{1+\theta \cos x}dx$

$\frac{\cos x}{1+\theta \cos x}=A+\frac{B}{1+\theta \cos x}$





$I'(\theta) = \int_0^{\pi}\frac{\cos x}{1+\theta \cos x}dx=\int_0^{\pi}A+\frac{B}{1+\theta \cos x}dx=\int_0^{\pi}\frac{1}{\theta}+\frac{\frac{-1}{\theta}}{1+\theta \cos x}dx$

$I'(\theta) = \frac{\pi}{\theta}-\frac{1}{\theta}\int_0^{\pi}\frac{1}{1+\theta \cos x}dx$

Do transform $\tan(x/2)=u$


$dx=\frac{2}{1+u^2} du$

I think after that you can handle the problem. If you cannot, please let me know

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.