Evaluate $$\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$$ where $a,b$ are finite natural numbers

I have spent about a day thinking over this problem. I tried integration by parts, differentiating under integral sign (Feynman's trick, with respect to $a, b$), using some trigonometric and logarithmic properties like changing $\cos^2\theta$ to $\cos2\theta$ and hereafter some logarithmic properties, etc., but failed miserably.

Also tried to use the property that $$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$ in-between, but still to no avail. I also tried to find similar questions on MSE but did not get a related one.

Can someone please help me to solve this integral?


My try (Feynman's trick) :

Let $$I(a)=\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) d\theta$$

Hence $$I'(a) =\frac 1a \int_0^{\pi/2} \frac {2a^2\cos^2\theta d\theta}{a^2\cos^2\theta +b^2\sin^2\theta}$$ $$=\frac 1a\left[ \frac {\pi}{2}+\int_0^{\pi/2} \frac {a^2\cos^2\theta -b^2\sin^2\theta}{a^2\cos^2\theta +b^2\sin^2\theta}\right]$$

Wherein between I broke $2a^2\cos^2\theta=a^2\cos^2\theta +b^2\sin^2\theta+a^2\cos^2\theta -b^2\sin^2\theta$

But now how do I continue further

  • $\begingroup$ Feynman's trick works like a charm here, can you show your attempt in full detail? $\endgroup$ Jun 12, 2018 at 17:06
  • $\begingroup$ I agree with Jack. Feynman's trick works here. $\endgroup$
    – Mark Viola
    Jun 12, 2018 at 17:06
  • 1
    $\begingroup$ "finite natural numbers" ? You can omit the "finite". $\endgroup$
    – Peter
    Jun 12, 2018 at 17:08
  • $\begingroup$ @Jack D'Aurizio I have edited my try accordingly. Can you please provide some hints further or any different parameter of differentiation in the same method $\endgroup$ Jun 12, 2018 at 17:15
  • 2
    $\begingroup$ The integral $$\int_{0}^{\pi/2}\frac{a^2\cos^2\theta-b^2\sin^2\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta$$ can be simply computed through the substitution $\theta=\arctan u$ and partial fraction decomposition. It equals $\frac{\pi}{2}\cdot\frac{a-b}{a+b}$, assuming that $a$ and $b$ have the same sign. $\endgroup$ Jun 12, 2018 at 17:15

4 Answers 4


Let $I(a,b)=\int_0^{\pi/2} \log(a^2\cos^2(\theta)+b^2\sin^2(\theta))\,d\theta$. Differentiating under the integral with respect to $a^2$ reveals

$$\begin{align} \frac{\partial I(a,b)}{\partial (a^2)}&=\int_0^{\pi/2}\frac{1}{a^2+b^2\tan^2(\theta)}\,d\theta\\\\ &=\frac{\pi/2}{a(a+b)}\tag1 \end{align}$$

Integrating $(1)$ with respect to $a^2$, we obtain

$$I(a,b)=\pi \log(a+b)+C$$

For $a=b$, $I(a,a)=\pi \log(a)$ from which we find that $C=-\pi\log(2)$.

Putting it all together yields

$$I(a,b)=\pi \log\left(\frac{a+b}2\right)$$


We see from symmetry that $I(a,b)=\pi \log\left(\frac{|a|+|b|}2\right)$ $\forall (a,b)\in \mathbb{R}^2\setminus (0,0)$.

  • $\begingroup$ Now both of you( Mark and Jack) have put in trouble. Both of your answers are marvelous in its own way. So which one to accept. $\endgroup$ Jun 12, 2018 at 17:28
  • $\begingroup$ Nope. Correction here. It is also applicable for $ab=0$ except for the case $a=b=0$ $\endgroup$ Jun 12, 2018 at 17:37
  • 1
    $\begingroup$ @Manthanein Good catch. I'll edit. $\endgroup$
    – Mark Viola
    Jun 12, 2018 at 17:41
  • $\begingroup$ Try $b=0$ and $a$ be some finite number. Now the integrals changes to $$\int_0^{\pi/2} \ln(a^2\cos^2\theta )d\theta=2\int_0^{\pi/2} \ln(a\cos\theta )d\theta=2\left[\frac {\pi}{2}\ln a-\frac {\pi}{2}\ln 2\right]=\pi\ln (a/2)$$ which matches with the general form you derived $\endgroup$ Jun 12, 2018 at 17:44
  • $\begingroup$ @Manthanein It should match, should it not? $\endgroup$
    – Mark Viola
    Jun 12, 2018 at 17:45

If we assume $a,b>0$ and set $$ I(a,b)=\int_{0}^{\pi/2}\log(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta $$ we have $I(a,b)=I(b,a)$ from the substitution $\theta\mapsto\frac{\pi}{2}-\theta$. On the other hand $I(a,a)=\pi\log(a)$ is trivial, so $I(a,b)=\pi\log\left(\frac{a+b}{2}\right)$ is a very reasonable conjecture. Indeed, it can be proved by computing $$ \frac{\partial I}{\partial a} = \int_{0}^{\pi/2}\frac{2a\cos^2\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta $$ as suggested in the comments, i.e. via $\theta\mapsto\arctan u$ and partial fraction decomposition.

An alternative approach is to notice that $$\begin{eqnarray*} I(a,b) &=& 2\,\text{Re}\int_{0}^{\pi/2}\log\left(a\cos\theta+ib\sin\theta\right)\\&=&\pi\log\left(\frac{a+b}{2}\right)+\text{Re}\int_{0}^{\pi}\log\left(e^{i\theta}+\frac{a-b}{a+b}\right)\,d\theta\end{eqnarray*}$$ where the last integral is a purely imaginary number by the residue theorem.

  • $\begingroup$ Now both of you( Mark and Jack) have put in trouble. Both of your answers are marvelous in its own way. So which one to accept. $\endgroup$ Jun 12, 2018 at 17:28
  • $\begingroup$ And by the way no need to assume $a, b\gt 0$, it's already provided in in question that they are natural numbers $\endgroup$ Jun 12, 2018 at 17:30
  • $\begingroup$ @Manthanein We need not assume that $a$ and $b$ are positive. Note that since the integral depends on $a^2$ and $b^2$, then the answer depends on $|a|$ and $|b|$ only. $\endgroup$
    – Mark Viola
    Jun 12, 2018 at 17:32
  • $\begingroup$ @Mark Viola That is what I said. Morever the modulus will be removed since they are natural and always positive $\endgroup$ Jun 12, 2018 at 17:34

Integrate by parts, then substitute $u=\tan\theta$ :

$$\begin{align*} I &= \int_0^{\tfrac\pi2} \ln\left(a^2\cos^2\theta + b^2\sin^2\theta\right) \, d\theta \\ &= \pi\ln a + \int_0^{\tfrac\pi2} \ln\left(1 + \left(\frac{b^2}{a^2}-1\right) \sin^2\theta\right) \, d\theta \\ &= \pi \ln b - 2\left(\frac{b^2}{a^2}-1\right) \int_0^{\tfrac\pi2} \frac{\theta \sin\theta \cos\theta}{1 + \left(\frac{b^2}{a^2}-1\right) \sin^2\theta} \, d\theta \\ &= \pi \ln b - 2\left(\frac{b^2}{a^2}-1\right) \int_0^\infty \frac{u \arctan u}{\left(u^2+1\right) \left(\frac{b^2}{a^2} u^2 + 1\right)} \, du \end{align*}$$

Now, for some real $\beta\neq0$, let

$$J(\alpha) = \int_0^\infty \frac{u \arctan(\alpha u)}{\left(u^2+1\right) \left(\beta^2u^2+1\right)} \, du \quad [I(0)=0]$$

Differentiate w.r.t. $\alpha$ and evaluate the subsequent integral. Partial fractions are our friend.

$$\frac{\partial J}{\partial \alpha} = \int_0^\infty \frac{u^2}{\left(u^2+1\right) \left(\beta^2u^2+1\right) \left(\alpha^2u^2+1\right)} = \frac\pi{2\left(\beta^2-1\right)} \left[\frac1{\alpha+1} - \frac1{\alpha+\beta}\right]$$

Integrate w.r.t. $\alpha$:

$$J(\alpha) = \frac\pi{2\left(\beta^2-1\right)} \int_0^\alpha \left(\frac1{t+1} - \frac1{t+\beta}\right) \, dt = \frac\pi{2\left(\beta^2-1\right)} \ln \frac{\beta(\alpha+1)}{\alpha+\beta}$$

Finally, replace $\beta=\dfrac ba$ and let $\alpha\to1$ to recover $I$:

$$J(1) = \frac{\pi a^2}{2\left(b^2-a^2\right)} \ln \frac{2b}{a+b} \implies \boxed{I = \pi \ln \frac{a+b}2}$$

  • 1
    $\begingroup$ That is quite a tedious way forward. But it does work. ;-) So (+1) $\endgroup$
    – Mark Viola
    Oct 19, 2023 at 21:21
  • $\begingroup$ If one really wants to get one's hands dirty, $J(1)$ can be written in terms of $J(a,b,c)$ as defined in this answer $\endgroup$
    – user170231
    Mar 22 at 1:00

Let $r= \frac{a-b}{a+b}$ \begin{align} & \int^{{\pi}/{2}}_{0}\ln{\left(a^2\sin^2t+b^2\cos^2t\right)}dt\\ = & \int^{\pi/2}_{0} \bigg[2\ln\frac{a+b}2+ \ln(1+r^2 -2r\cos 2t)\bigg] dt = \pi\ln\frac{a+b}2 \end{align} where \begin{align} & \int^{{\pi}/{2}}_{0} \ln(1+r^2 -2r\cos 2t) dt\\ = & \int^{\pi/2}_{0} \int_0^{r} \frac{2s-2\cos2t}{1+s^2 -2s\cos 2t}ds \ dt\\ = &\ \int_0^{r} \left(\frac1s\tan^{-1} \frac{s\sin2t}{s\cos 2t-1}\right)_{t=0}^{t=\frac\pi2}ds = \int_0^r 0 \ ds=0 \end{align}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .