Question

The value of $\int_{0}^{1}\frac{ln(x+x^{-1})}{x+x^{-1}}dx$ equals (It is given that $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}$)

Answer 1

$\frac{(\ln 2)^2}{4}-\frac{\pi^2}{48}$

Answer 2

The given integral is $I = \int_{0}^{1}\frac{\ln(x+x^{-1})}{x+x^{-1}}dx$.

First, simplify the term $x+x^{-1} = x+\frac{1}{x} = \frac{x^2+1}{x}$. So the integral becomes $I = \int_{0}^{1}\frac{\ln\left(\frac{x^2+1}{x}\right)}{\frac{x^2+1}{x}}dx$.

Let's use the substitution $x = \tan\theta$. When $x=0$, $\theta=0$. When $x=1$, $\theta=\frac{\pi}{4}$. $dx = \sec^2\theta d\theta$. The term $x+x^{-1}$ becomes $\tan\theta + \frac{1}{\tan\theta} = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} = \frac{2}{2\sin\theta\cos\theta} = \frac{2}{\sin(2\theta)}$.

Substitute these into the integral: $I = \int_{0}^{\pi/4} \frac{\ln\left(\frac{2}{\sin(2\theta)}\right)}{\frac{2}{\sin(2\theta)}} \sec^2\theta d\theta$ $I = \int_{0}^{\pi/4} \frac{\sin(2\theta)}{2} \ln\left(\frac{2}{\sin(2\theta)}\right) \sec^2\theta d\theta$ Using $\sin(2\theta) = 2\sin\theta\cos\theta$ and $\sec^2\theta = \frac{1}{\cos^2\theta}$: $I = \int_{0}^{\pi/4} \frac{2\sin\theta\cos\theta}{2} \left(\ln 2 - \ln(\sin(2\theta))\right) \frac{1}{\cos^2\theta} d\theta$ $I = \int_{0}^{\pi/4} \frac{\sin\theta}{\cos\theta} \left(\ln 2 - \ln(\sin(2\theta))\right) d\theta$ $I = \int_{0}^{\pi/4} \tan\theta (\ln 2 - \ln(\sin(2\theta))) d\theta$ Split the integral into two parts: $I = \ln 2 \int_{0}^{\pi/4} \tan\theta d\theta - \int_{0}^{\pi/4} \tan\theta \ln(\sin(2\theta)) d\theta$

Evaluate the first integral: $\int_{0}^{\pi/4} \tan\theta d\theta = [\ln|\sec\theta|]_{0}^{\pi/4} = \ln(\sec(\pi/4)) - \ln(\sec(0)) = \ln(\sqrt{2}) - \ln(1) = \frac{1}{2}\ln 2 - 0 = \frac{1}{2}\ln 2$ So the first term is $\ln 2 \cdot \frac{1}{2}\ln 2 = \frac{1}{2}(\ln 2)^2$.

Now, let $J = \int_{0}^{\pi/4} \tan\theta \ln(\sin(2\theta)) d\theta$. Apply integration by parts: $\int u dv = uv - \int v du$. Let $u = \ln(\sin(2\theta))$ and $dv = \tan\theta d\theta$. Then $du = \frac{1}{\sin(2\theta)} \cdot 2\cos(2\theta) d\theta = 2\cot(2\theta) d\theta$. And $v = \int \tan\theta d\theta = \ln|\sec\theta|$.

$J = [\ln|\sec\theta|\ln(\sin(2\theta))]_{0}^{\pi/4} - \int_{0}^{\pi/4} \ln|\sec\theta| \cdot 2\cot(2\theta) d\theta$ Evaluate the boundary term: At $\theta=\frac{\pi}{4}$: $\ln(\sec(\pi/4))\ln(\sin(\pi/2)) = \ln(\sqrt{2})\ln(1) = \frac{1}{2}\ln 2 \cdot 0 = 0$. At $\theta=0$: $\lim_{\theta \to 0^+} \ln(\sec\theta)\ln(\sin(2\theta))$. This is a $0 \times (-\infty)$ form. Using L'Hopital's rule or series expansion: $\ln(\sec\theta) = -\ln(\cos\theta) \approx -(\frac{\theta^2}{2}-1) \approx \frac{\theta^2}{2}$ for small $\theta$. $\ln(\sin(2\theta)) \approx \ln(2\theta)$ for small $\theta$. So the limit is $\lim_{\theta \to 0^+} \frac{\theta^2}{2} \ln(2\theta)$. Let $y=2\theta$. $\lim_{y \to 0^+} \frac{y^2}{8}\ln y = 0$. Thus, the boundary term is $0$.

So, $J = - \int_{0}^{\pi/4} \ln|\sec\theta| \cdot 2\cot(2\theta) d\theta = \int_{0}^{\pi/4} \ln(\cos\theta) \cdot 2\cot(2\theta) d\theta$. We have $2\cot(2\theta) = \frac{2\cos(2\theta)}{\sin(2\theta)} = \frac{2(2\cos^2\theta-1)}{2\sin\theta\cos\theta} = \frac{2\cos^2\theta-1}{\sin\theta\cos\theta} = \frac{2\cos\theta}{\sin\theta} - \frac{1}{\sin\theta\cos\theta} = 2\cot\theta - 2\csc(2\theta)$. So, $J = \int_{0}^{\pi/4} \ln(\cos\theta) (2\cot\theta - 2\csc(2\theta)) d\theta$. This approach seems to lead to more complex integrals.

Let's use a different substitution in the original integral. $I = \int_{0}^{1}\frac{\ln(x+x^{-1})}{x+x^{-1}}dx$. Let $u = x+\frac{1}{x}$. Then $du = (1-\frac{1}{x^2})dx = \frac{x^2-1}{x^2}dx$. This doesn't seem to simplify the integral directly.

Consider the property of definite integrals: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here $a=0, b=1$. So $x \to 1-x$. $I = \int_{0}^{1}\frac{\ln((1-x)+(1-x)^{-1})}{(1-x)+(1-x)^{-1}}dx$. This makes it more complicated.

Let's try a substitution $x=e^{-t}$. When $x=0$, $t \to \infty$. When $x=1$, $t=0$. $dx = -e^{-t} dt$. $x+x^{-1} = e^{-t}+e^t = 2\cosh t$. $I = \int_{\infty}^{0} \frac{\ln(2\cosh t)}{2\cosh t} (-e^{-t}) dt$ $I = \int_{0}^{\infty} \frac{\ln(2\cosh t)}{2\cosh t} e^{-t} dt$ Expand $2\cosh t = e^t+e^{-t}$: $I = \int_{0}^{\infty} \frac{\ln(e^t+e^{-t})}{e^t+e^{-t}} e^{-t} dt$ $I = \int_{0}^{\infty} \frac{\ln(e^t(1+e^{-2t}))}{e^t(1+e^{-2t})} e^{-t} dt$ $I = \int_{0}^{\infty} \frac{t+\ln(1+e^{-2t})}{e^t+e^{-t}} e^{-t} dt$ $I = \int_{0}^{\infty} \frac{t e^{-t}}{e^t+e^{-t}} dt + \int_{0}^{\infty} \frac{\ln(1+e^{-2t})e^{-t}}{e^t+e^{-t}} dt$ Multiply numerator and denominator by $e^{-t}$: $I = \int_{0}^{\infty} \frac{t e^{-2t}}{1+e^{-2t}} dt + \int_{0}^{\infty} \frac{\ln(1+e^{-2t})e^{-2t}}{1+e^{-2t}} dt$ Let $u = e^{-2t}$. Then $du = -2e^{-2t} dt$. So $e^{-2t} dt = -\frac{1}{2}du$. Also, $t = -\frac{1}{2}\ln u$. When $t=0$, $u=e^0=1$. When $t=\infty$, $u=e^{-\infty}=0$.

Substitute these into the first integral: $I_1 = \int_{1}^{0} \frac{-\frac{1}{2}\ln u}{1+u} \left(-\frac{1}{2}du\right) = \frac{1}{4} \int_{0}^{1} \frac{\ln u}{1+u} du$ This is a standard integral. We know that $\frac{1}{1+u} = \sum_{n=0}^{\infty} (-1)^n u^n$ for $|u|<1$. So, $\int_{0}^{1} \frac{\ln u}{1+u} du = \int_{0}^{1} \ln u \sum_{n=0}^{\infty} (-1)^n u^n du = \sum_{n=0}^{\infty} (-1)^n \int_{0}^{1} u^n \ln u du$. The integral $\int_{0}^{1} u^n \ln u du$ can be evaluated by integration by parts: $\int u^n \ln u du = \frac{u^{n+1}}{n+1}\ln u - \int \frac{u^{n+1}}{n+1}\frac{1}{u} du = \frac{u^{n+1}}{n+1}\ln u - \frac{u^{n+1}}{(n+1)^2}$. Evaluating from $0$ to $1$: $[\frac{u^{n+1}}{n+1}\ln u - \frac{u^{n+1}}{(n+1)^2}]_{0}^{1} = (0 - \frac{1}{(n+1)^2}) - (0 - 0) = -\frac{1}{(n+1)^2}$. (The limit $\lim_{u\to 0^+} u^{n+1}\ln u = 0$ for $n+1>0$.) So, $\int_{0}^{1} \frac{\ln u}{1+u} du = \sum_{n=0}^{\infty} (-1)^n \left(-\frac{1}{(n+1)^2}\right) = -\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2}$. Let $k=n+1$. Then $n=k-1$. When $n=0, k=1$. When $n=\infty, k=\infty$. So, $-\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2} = -\sum_{k=1}^{\infty} \frac{-(-1)^k}{k^2} = \sum_{k=1}^{\infty} \frac{(-1)^k}{k^2}$. This is related to the given sum $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}$. Note that $\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} = \frac{\pi^2}{12}$. Our sum is $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2} = - \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} = -\frac{\pi^2}{12}$. Therefore, $I_1 = \frac{1}{4} \left(-\frac{\pi^2}{12}\right) = -\frac{\pi^2}{48}$.

Now, substitute $u=e^{-2t}$ into the second integral: $I_2 = \int_{1}^{0} \frac{\ln(1+u)}{1+u} \left(-\frac{1}{2}du\right) = \frac{1}{2} \int_{0}^{1} \frac{\ln(1+u)}{1+u} du$ This integral is of the form $\int \frac{\ln y}{y} dy = \frac{(\ln y)^2}{2}$. Let $y = 1+u$. Then $dy = du$. When $u=0, y=1$. When $u=1, y=2$. $I_2 = \frac{1}{2} \int_{1}^{2} \frac{\ln y}{y} dy = \frac{1}{2} \left[\frac{(\ln y)^2}{2}\right]_{1}^{2}$ $I_2 = \frac{1}{4} [(\ln 2)^2 - (\ln 1)^2] = \frac{1}{4} [(\ln 2)^2 - 0] = \frac{(\ln 2)^2}{4}$

Finally, $I = I_1 + I_2 = -\frac{\pi^2}{48} + \frac{(\ln 2)^2}{4}$.