This problem has been troubling me for a while now; even though I have tried my best ( and filled up a rough notebook in the process). Consider $I_1=\int_{0}^{1} \frac{\tan^{-1}x}{x} dx$, and $I_2=\int_{0}^{\pi/2} \frac{x}{\sin x}dx$. We are supposed to find $\frac{I_1}{I_2}$. The answer is $1/2$.

My try- To make the limits for both identical, I substituted $x=\sin\theta$ in the first integral, and then tried to make use of the properties of definite integrals ( replacing $f(\theta)$ by $f(\pi/2-\theta)$ etc), but no real progress was made. I then tried $x=\arcsin t$ for the second one, but no result.

Now I really doubt if there is something wrong with the question itself, or am I just being really silly. Please help me here. Thanks in advance!!

up vote 1 down vote accepted

If you set $t=2\arctan x$ in the first integral, you have $x=\tan(t/2)$, so $$ dx= \frac{1}{2}\left(1+\tan^2\frac{t}{2}\right)dt= \frac{1}{2}\left(1+\frac{1-\cos t}{1+\cos t}\right)dt= \frac{1}{1+\cos t}\,dt $$ Also recall that $$ \tan\frac{t}{2}=\frac{\sin t}{1+\cos t} $$

With $x=\tan t$ then $u=2t$,

$$\int_0^1\frac{\arctan x}xdx=\int_0^{\pi/4}\frac{t\,dt}{\sin t\cos t}=\frac12\int_0^{\pi/2}\frac{u\,du}{\sin u}.$$

While the previous two answers are undoubtedly the most efficient ways to answer the question, I was curious about the two integrals, so I decided to evaluate them.

For the first integral $$\int\limits_{0}^{1} \frac{\tan^{-1}(x)}{x} \mathrm{d}x$$ expand the inverse tangent in a Maclaurin series \begin{equation} \tan^{-1}(x) = \sum\limits_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{2n+1} \end{equation} divide by $x$ and integrate term-by-term.

Thus, we have \begin{equation} \int\limits_{0}^{1} \frac{\tan^{-1}(x)}{x} \mathrm{d}x = \sum\limits_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} = \mathrm{G} \end{equation} where G is Catalan's constant.

For the second integral $$\int\limits_{0}^{\pi /2} \frac{x}{\sin(x)} \mathrm{d}x$$ We begin by expanding the denominator in exponential functions and evaluating the following indefinite integral \begin{align} \int \frac{1}{\mathrm{e}^{ix}-\mathrm{e}^{-ix}} \mathrm{d}x &= -\int \frac{\mathrm{e}^{ix}}{1-\mathrm{e}^{i2x}} \mathrm{d}x \\ &= -\int \frac{\mathrm{e}^{ix}}{(1+\mathrm{e}^{ix})(1-\mathrm{e}^{ix})} \mathrm{d}x \\ &= -\frac{1}{2} \int \Big[\frac{\mathrm{e}^{ix}}{1+\mathrm{e}^{ix}} + \frac{\mathrm{e}^{ix}}{1-\mathrm{e}^{ix}} \Big] \mathrm{d}x \\ &= \frac{i}{2} \big[\ln(1+\mathrm{e}^{ix}) - \ln(1-\mathrm{e}^{ix})\big] + \mathrm{const} \tag{1} \label{eq:xosinx-1} \end{align}

Now we evaluate the following integral by parts \begin{equation} \int \frac{x}{\mathrm{e}^{ix}-\mathrm{e}^{-ix}} \mathrm{d}x = ab \,- \int b \,\mathrm{d}a \tag{2} \label{eq:xosinx-2} \end{equation} where $a = x ,\, \mathrm{d}a = \mathrm{d}x ,\, \mathrm{d}b = \mathrm{d}x/(\mathrm{e}^{ix}-\mathrm{e}^{-ix}) ,\, b = $ equation \eqref{eq:xosinx-1}.

The integral on the right in equation \eqref{eq:xosinx-2} is \begin{align} \int \ln(1+\mathrm{e}^{ix}) \mathrm{d}x &= i \int \frac{\ln(u)}{1-u} \mathrm{d}u \\ &= -i \int \frac{\ln(1-y)}{y} \mathrm{d}y \\ &= i \mathrm{Li}_{2}(y) \\ & = i \mathrm{Li}_{2}(-\mathrm{e}^{ix}) \end{align} where we used the following substitutions in succession: $u = 1 + \mathrm{e}^{ix} ,\, y = 1-u$. $\mathrm{Li}_{2}(z)$ is the dilogarithm.

Likewise, we have \begin{equation} \int \ln(1-\mathrm{e}^{ix}) \mathrm{d}x = i \mathrm{Li}_{2}(\mathrm{e}^{ix}) \end{equation}

Putting all of the pieces together, we obtain \begin{align} \int\limits_{0}^{\pi /2} \frac{x}{\sin(x)} \mathrm{d}x &= i2 \int\limits_{0}^{\pi /2} \frac{x}{\mathrm{e}^{ix}-\mathrm{e}^{-ix}} \mathrm{d}x \\ &= x\big[\ln(1-\mathrm{e}^{ix}) - \ln(1+\mathrm{e}^{ix}) \big] + i\Big[\mathrm{Li}_{2}(-\mathrm{e}^{ix}) - \mathrm{Li}_{2}(\mathrm{e}^{ix}) \Big] \Big|_{0}^{\pi /2} \\ &= \frac{\pi}{2} \big[\ln(1-i) - \ln(1+i) \big] + i\Big[\mathrm{Li}_{2}(-i) - \mathrm{Li}_{2}(i)\Big] - 0 -i\Big[\mathrm{Li}_{2}(-1) - \mathrm{Li}_{2}(1)\Big] \\ &= \frac{\pi}{2} \Big[\frac{\ln(2)}{2} -i\frac{\pi}{4} - \frac{\ln(2)}{2} - i\frac{\pi}{4}\Big] + i\Big[-\frac{\pi^{2}}{48}-i\mathrm{G}+\frac{\pi^{2}}{48}-i\mathrm{G} \Big] - i\Big[-\frac{\pi^{2}}{12}-\frac{\pi^{2}}{6} \Big] \\ &= 2\mathrm{G} \end{align}

  • 1
    Sort of amazing how all those terms cancel out. – marty cohen Sep 18 '16 at 1:55
  • impressive work +1 – Lelouch.D.Light May 19 '17 at 13:29

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