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This equality showed up in the process of computing the integral $I$ below with the residue theorem: $$I = \int_0^\tfrac\pi2 \frac x{\cos x+\sin x}\,dx = \int_1^\infty \frac{x\log x}{\left(1+x^2\right)\sqrt{x^2-1}}\,dx = J.$$

The nearest I can get $J$ to resemble $I$ is $$\begin{align*}J &= \int_0^\tfrac\pi2 \frac{\log(\csc y)}{1+\sin^2y} \, dy & x=\csc y \\ &= \int_0^\tfrac\pi2 \frac{\arctan\left(\sqrt2\,\tan y\right)}{\sqrt2\,\tan y} \, dy & \text{by parts} \\ &= \int_0^\tfrac\pi2 \frac{\sqrt2\,z}{(1+\cos^2z)\tan z} \, dz & \tan z=\sqrt2\,\tan y. \end{align*}$$

At this point I gave up and resorted to the Feynman approach to calculate $J$ using the $\arctan$ form above. Is there a simpler method, e.g. substitution or other observation that can be made, to arrive at $I\equiv J$?

In an attempt to prove that $I-J=0$, I joined the integrands, substituted $x=\tan\dfrac y2$, and decomposed into partial fractions to yield

$$\sqrt2 \int_0^1 \left(\frac{2y^3}{1+y^4}+\frac{2\sqrt2}{1+2y-y^2}-\frac1y\right) \arctan y \, dy = 0,$$

the last term of which evaluates to $G$ (Catalan's constant). Another integral for $G$ then follows:

$$\int_0^1 \left(\frac{y^3}{1+y^4}+\frac{\sqrt2}{1+2y-y^2}\right) \arctan y \, dy = \frac G2$$

but I have no productive ideas on how to establish this.

Another thing I've considered in the vein of proving $I-J=0$ was to instead substitute $x=\tan y$ into both, then reduce as much as possible. I admit, resorting to contour integration again is self-defeating as far as finding a "simple" connection between $I$ and $J$ is concerned. But at the very least, this produces a much cleaner vanishing integral.

$$\begin{align*} I &= \int_0^\infty \frac{\arctan y}{(1+y)\sqrt{1+y^2}} \, dy & y=\tan x\\ I' &= \int_0^\infty \frac{\frac\pi2-\arctan y}{(1+y)\sqrt{1+y^2}} \, dy & y\to\frac 1y \\ I = \frac{I+I'}2 &= \frac\pi4 \int_0^\infty \frac{dy}{(1+y)\sqrt{1+y^2}} \\ &= \frac\pi{2\sqrt2} \int_0^1 \frac{dz}{\sqrt{1+z^2}} & y=\frac{1-z}{1+z};\text{ symmetry} \\ &= \frac\pi{2\sqrt2} \int_0^\tfrac1{\sqrt2} \frac{dw}{1-w^2} & w=\frac z{\sqrt{1+z^2}} \\ &= \frac\pi2 \int_0^1 \frac{dw}{2-w^2} & w\to\frac w{\sqrt2} \\[2ex] \hline J &= \sqrt2 \int_0^\infty \frac{\arctan y}{y\left(2+y^2\right)} \, dy & y=\tan x \\ &= \frac1{\sqrt2} \int_0^1 \int_{-\infty}^\infty \frac{dy\,dz}{\left(1+y^2z^2\right)\left(2+y^2\right)} & \text{Fubini; symmetry} \\ &= \frac\pi2 \int_0^1 \frac{dz}{1+\sqrt2\,z} & \text{residue thm. (semicircle)} \\[2ex] \hline \implies I-J &= \frac\pi2 \int_0^1 \left(\frac1{2-u^2} - \frac1{1+\sqrt2\,u}\right) \, du & (\star) \\ &= \frac\pi2 \int_0^1 \frac{-1+\sqrt2\,u+u^2}{\left(1+\sqrt2\,u\right)\left(2-u^2\right)} \, du \end{align*}$$

The integral $(\star)$ is very easy to evaluate with the fundamental theorem of calculus. Is there a more efficient way of recovering it from $I-J$ than the above? And is there a quick(er) way of deducing that it's identically zero?

The plot of the $u$-integrand almost suggests symmetry about $u=\dfrac12$, but that's unfortunately not the case since the numerator actually has a root at $u=\dfrac{\sqrt6-\sqrt2}2\approx0.518$.

user170231
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    Symmetry does not seem to work in this case. Perhaps the substitution $x = \textrm{arctan} \left(\sqrt{2} \tan y\right)$ would work better. – Bumblebee Jun 04 '25 at 03:20
  • Simpler than with Feynman's trick, I do not think. I found several methods for computing the antiderivatives and the definite integrals but they all involve a buch of polylogarithms and logarithms of complex arguments. – Claude Leibovici Jun 04 '25 at 03:41
  • $J=\sqrt2\int_0^\infty \frac{\arctan(t)}{t(t^2+2)},dt$ $J(a)=\int_0^\infty \frac{\arctan(at)}{t(t^2+2)},dt\underset{I(0)=0}{\implies }J'(a)=\int_0^\infty \frac{1}{(t^2+2)(1+a^2t^2)},dt=\frac{\pi}{4a + \sqrt[2]{8}}$

    $J(a)=\int\frac{\pi}{4a + \sqrt[2]{8}},da=\frac{\pi \ln\left(\left|4a + 2^{\frac{3}{2}}\right|\right)}{4}+C$

    $\therefore J(a)=\frac{\pi \ln\left(\left|4a + 2^{\frac{3}{2}}\right|\right)}{4}-\frac{\pi}{4}\ln(\sqrt{8})\implies \sqrt2J(1)=\color{red}{\frac{\sqrt2\pi \ln\left(\left|4 + 2^{\frac{3}{2}}\right|\right)}{4}-\frac{\sqrt2\pi}{4}\ln(\sqrt{8})}$

     – Amrut Ayan Jun 04 '25 at 05:05
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    I can prove it by evaluating both integrals, which is a bad way – Martin.s Jun 04 '25 at 08:53
  • @AmrutAyan which can be simplified as $J(1)=\frac\pi{2\sqrt2}\sinh^{-1}1$. – Integreek Jun 04 '25 at 10:35
  • @Martin.s Why is a trivial evaluation a "bad way"? – user Jun 04 '25 at 10:42
  • @Integreek thanks(hadn’t simplified my result at all) – Amrut Ayan Jun 04 '25 at 12:19
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    @user I think Martin is acknowledging that evaluating $I$ and $J$ separately is outside the scope of this question. We know $I=J$, but I'm hoping there is a way to connect them without having to completely and/or explicitly evaluate them, much like was done in this question. – user170231 Jun 04 '25 at 13:38
  • Just to clarify : may we compute the antiderivatives or not ? – Claude Leibovici Jun 06 '25 at 12:12
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    @ClaudeLeibovici I'd prefer to see a manipulation of the definite integral if possible, i.e. converting $I$ to $J$ or vice versa, or proving that the integral $I-J$ is vanishing. That said, I've seen the antiderivatives that Mathematica has to offer. I can use its output to reduce $I$ to the known closed form, but $J$ is still evading me as it requires simplification of this dilog sum – user170231 Jun 06 '25 at 13:35
  • I have not be able to do anything else – Claude Leibovici Jun 06 '25 at 16:42

1 Answers1

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$$\large\color{blue}{J=\int\frac{\sqrt2\,x}{(1+\cos^2(x))\tan (x)}\,dx}$$ One integration by part gives $$J=\frac x {2\sqrt 2}\log \left(\frac{\sin ^2(x)}{2-\sin ^2(x)}\right)-\frac 1 {2\sqrt 2}\int \log \left(\frac{\sin ^2(x)}{2-\sin ^2(x)}\right)\,dx$$

Using the bounds, the first term is $0$.

$$K=\int \log \left(\frac{\sin ^2(x)}{2-\sin^2(x)}\right)\,dx=\int \log \left(\frac{1-\cos (2 x)}{3+\cos (2 x)}\right)\,dx$$ $$x=\tan^{-1}(t) \quad \implies \quad K=\int \frac{1}{t^2+1}\log \left(\frac{t^2}{t^2+2}\right)\,dt$$ Write $$\log \left(\frac{t^2}{t^2+2}\right)=2\log(t)-\log \left(t-i \sqrt{2}\right)-\log\left(t+i \sqrt{2}\right)$$ and $$\frac{1}{t^2+1}=\frac i 2 \Big( \frac{1}{t+i}-\frac{1}{t-i} \Big)$$ This leads to "simple" integrals (alas a bunch of logarithms and polylogarithms with complex arguments).

Back to $x$ and using the bounds $$K=-\frac{1}{2} \pi \log \left(3+2\sqrt{2}\right)\quad \implies \quad J=\frac{\pi }{4 \sqrt{2}}\log \left(3+2\sqrt{2}\right)$$

$$\large\color{blue}{I=\int\frac x{\cos (x)+\sin (x)}\,dx}$$

One integration by parts gives $$I=\sqrt{2} x \tanh ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)-1}{\sqrt{2}}\right)-\sqrt{2}\int \tanh ^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)-1}{\sqrt{2}}\right)\,dx$$ Using the bounds, the first term is $0$.

$$L=\int \tanh ^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)-1}{\sqrt{2}}\right)\,dx$$ Using the logarithmic form of the hyperbolic arc tangent function leads again to logarithms and polylogarithms with complex arguments. Using the bounds $$L=\frac{\pi}{8} \log \left(3-2 \sqrt{2}\right)\quad \implies \quad I=-\frac{\pi }{4 \sqrt{2}} \log \left(3-2 \sqrt{2}\right)=\frac{\pi }{4 \sqrt{2}}\log \left(3+2\sqrt{2}\right)$$

Claude Leibovici
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