Compute the following improper integral with parameters:

Question

For $a,b>0$, evaluate $$I(a,b)= \int_0^\infty \ln(a^2\sin^2x+b^2\cos^2x)dx.$$

What I thought of is that I can try to find an integral $\int_a^b \text{(something)}dy$ such that the result would be $\ln(a^2\sin^2x+b^2\cos^2x)$ (and therefore obtaining a double integral that can be easily computed). Is that the correct approach?
P.S. : The idea that stood behind this incomplete reasoning was the striking similarity between the formula $\int \frac{x}{x^2+a^2}=\frac{1}{2}\ln(x^2+a^2)+C$ and what I have above.

Please check the upper limit... as written, your integral is not convergent. — Olivier Oloa, Apr 08 '18 at 16:30
However, a similar problem might be the integral$$\int\limits_{-\pi/2}^{\pi/2}dx,\log(a^2\cos^2x+b^2\sin^2x)=2\pi\log\left(\frac {a+b}2\right)$$which is a classic problem — Crescendo, Apr 08 '18 at 16:53

Robert Z · Accepted Answer · 2018-04-08T17:29:11.897

2

The function $f(x)=\ln(a^2\sin^2x+b^2\cos^2x)$ is periodic of period $\pi$ (we assume that $a$ and $b$ are such that $a^2\sin^2x+b^2\cos^2x>0$). Therefore, for $t>0$, $$\int_0^{t} f(x)dx=n\int_0^{\pi} f(x)dx+\int_0^{t-n\pi} f(x)dx$$ where $n=\lfloor t/\pi\rfloor$.

We claim that the improper integral on $[0,+\infty)$ is convergent iff $f$ is identically zero that is when $a^2=b^2=1$.

It diverges when $\int_0^{\pi} f(x)dx\not=0$. If $\int_0^{\pi} f(x)dx=0$ then the integral function $$t\to \int_0^{t} f(x)dx=\int_0^{t-n\pi} f(x)dx$$ is periodic (and not identically zero) and therefore the limit $\lim_{t\to +\infty}\int_0^{t} f(x)dx$ does not exist.

edited Apr 08 '18 at 17:29

answered Apr 08 '18 at 16:30

Robert Z

145,942

$a^2\sin^2(x)+b^2\cos^2(x)$ can certainly be smaller than $1$ for some values of $x\in [0,\pi]$ and exceed $1$ for others. So, how does one know that the integral diverges for all $a$ an $b$? There might be a choice of $a$ and $b$ such that $\int_0^\pi f(x),dx=0$. – Mark Viola Apr 08 '18 at 16:50
@Mark Viola I claim that it diverges if $\int_0^{\pi} f(x)dx\not=0$, otherwise the limit of $\int_0^{t} f(x)dx$ as $t\to +\infty$ does not exist. – Robert Z Apr 08 '18 at 16:54
And why does it fail to exist if $\int_0^\pi f(x),dx=0$? That is the important part of the proof here. – Mark Viola Apr 08 '18 at 16:56
Because $t\to \int_0^{t} f(x)dx=\int_0^{t-n\pi} f(x)dx$ is periodic (and not identically zero). – Robert Z Apr 08 '18 at 16:58
Yes, and I am suggesting your adding that to your answer. – Mark Viola Apr 08 '18 at 16:59
@MarkViola Done! Thank you for pointing out. – Robert Z Apr 08 '18 at 17:08
(+1) And Robert, for $a=b=1$, the integral does exist, which you implied by rejecting the trivial case. – Mark Viola Apr 08 '18 at 17:20
@MarkViola Yes, you are right. I revised my answer. Thanks again. – Robert Z Apr 08 '18 at 17:30
Pleased I was able to provide input. – Mark Viola Apr 08 '18 at 18:10

Compute the following improper integral with parameters:

1 Answers1