Evaluate $\int_0^{\pi/2}\log(a^2\sin^2x+b^2\cos^2x)dx$

calculusintegrationtrigonometric-integrals

Evaluate $$\int_0^{\pi/2}\log(a^2\sin^2x+b^2\cos^2x)dx$$
Basically, I have tried all the definite integrals formula I know, but nothing works. Please help me. I can't figure out actually how to handle this one.

Thanks in advance.

Best Answer

Use the known result $\int^{\pi}_{0} {\ln(1+r^2 -2r\cos t)}dt=0$ to obtain

\begin{align} & \int^{\frac{\pi}{2}}_{0}{\ln{\left(a^2\sin^2x+b^2\cos^2x\right)}dx}\\ = & \frac12\int^{\pi}_{0}{\ln{\left(\frac{a^2+b^2}2-\frac{a^2-b^2}2\cos t\right)}dt}\\ = & \frac12\int^{\pi}_{0} {\ln\left( \left(\frac{|a|+|b|}2\right)^2\cdot (1+r^2 -2r\cos t)\right)}dt ,\>\>\> r=\frac{|a|-|b|}{|a|+|b|}\\ = & \pi\ln\frac{|a|+|b|}2 \end{align}

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[Math] Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$

Here's one way to go.

First, note that $$\begin{eqnarray*} \int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx &=& \int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx +\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx. \end{eqnarray*}$$ For now I'll simply claim that \begin{equation*} \int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx = \pi\log 64.\tag{1} \end{equation*} (I would be surprised if this integral has not been handled somewhere on this site.) But $$\begin{eqnarray*} \int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx &=& \int_0^\pi\frac{3x}{\sin x} \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k}} \sin^{2k}x \ \mathrm dx \\ &=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}} \int_0^\pi x \sin^{2k-1}x \ \mathrm dx \\ &=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}} \frac{\pi^{3/2}\Gamma(k)}{2\Gamma(k+1/2)} \\ &=& -\pi \sum_{k=1}^\infty \frac{1}{3^{2k-1}2k(2k-1)} \\ &=& -\pi \log \frac{32}{27}. \end{eqnarray*}$$ (The last sum can be found by standard methods. Schematically, $\sum \frac{a^{2k-1}}{2k(2k-1)} = \sum \int {\mathrm da} \frac{a^{2k-2}}{2k}$.) Thus, the integral is $\pi \log 54$ as claimed.

Proof of (1): We have $$\begin{eqnarray*} \int_0^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx &=& \int_{0^+}^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx \\ &=& 3\int_{0^+}^\pi x \cot\frac{x}{2} \ \mathrm dx \hspace{5ex}\textrm{(double angle formulas)} \\ &=& 12 \int_{0^+}^{\pi/2} t\cot t \ \mathrm dt \hspace{5ex} (t = x/2) \\ &=& -12\int_{0^+}^{\pi/2} \log\sin t \ \mathrm dt \hspace{5ex}\textrm{(integrate by parts)} \\ &=& -6\int_{0^+}^{\pi/2} \log\sin^2 t \ \mathrm dt \\ &=& -6\int_{0^+}^{\pi/2} \log(1-\cos^2 t) \ \mathrm dt \\ &=& 6 \int_{0^+}^{\pi/2} \sum_{k=1}^\infty \frac{1}{k}\cos^{2k}t \ \mathrm dt \hspace{5ex}\textrm{(series for log)} \\ &=& 6\sum_{k=1}^\infty \frac{1}{k} \int_{0^+}^{\pi/2} \cos^{2k}t \ \mathrm dt \hspace{5ex} \textrm{(Tonelli's theorem)}\\ &=& 6\sum_{k=1}^\infty \frac{1}{k} \frac{\sqrt{\pi}\Gamma(k+1/2)}{2\Gamma(k+1)} \\ &=& 3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1}\frac{2k-1}{k} \\ &=& \pi \log 64. \end{eqnarray*}$$ Note that $$\begin{eqnarray*} 6\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} &=& -6\pi \left[\sum_{k=0}^\infty {1/2 \choose k}(-1)^{k} - 1\right] \\ &=& -6\pi[(1-1)^{1/2} - 1] \\ &=& 6\pi \end{eqnarray*}$$ and $$\begin{eqnarray*} -3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} \frac{1}{k} &=& 3\pi \sum_{k=1}^\infty {1/2\choose k}(-1)^k \int_0^1 x^{k-1} \ \mathrm dx \\ &=& 3\pi \int_0^1 \frac{1}{x} \left[ \sum_{k=0}^\infty {1/2\choose k}(-1)^k x^{k} -1 \right] \ \mathrm dx \\ &=& 3\pi \int_0^1 \frac{1}{x} \left( \sqrt{1-x} -1 \right) \ \mathrm dx \\ &=& 3\pi(-2+\log 4) \\ &=& -6\pi + \pi\log 64. \end{eqnarray*}$$

[Math] Find $\lim_{n \to \infty} n \int_0^1 (\cos x – \sin x)^n dx$

Write the integrand as $(\cos(x)-\sin(x))^n=e^{n\log(\cos(x)-\sin(x))}$. Looking at a few plots it becomes immediatly clear the integral will be dominated by a small region around the origin with width $\epsilon\sim1/n$ as $n \rightarrow\infty$. The more formal reason for that is, that the exponent is nearly zero around the origin

$$ I_n=\int_0^1(\cos(x)-\sin(x))^n\sim \int_0^{\epsilon}e^{n\log(\cos(x)-\sin(x))} $$

Taylor expansion of the exponent yields

$$ I_n\sim\int_0^{\epsilon}e^{-x n} $$

by pushing $\epsilon$ to infinity we introduce only an exponentially small error so

$$ I_n \sim\int_0^{\infty}e^{-x n}=\frac{1}{n} $$

which yields

$$ \lim_{n \rightarrow \infty} n I_n=1 $$

Best Answer

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[Math] Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$

[Math] Find $\lim_{n \to \infty} n \int_0^1 (\cos x – \sin x)^n dx$

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