After weeks of going back and forth I've been able to solve the following definite integral:

$$I = \int_{0}^{\frac{\pi}{2}} \ln\left|\sec^2(x) + \tan^4(x) \right|\:dx $$

To solve this I employ Feynman's Trick with Glasser's Master Theorom but I'm excited to learn of other methods that can be employed. Are there any other 'tricks' that can be used? or alternatively series based solutions? or transformations? (or anything for that matter).

For those who may be interested my process was:

(1) First make the substitution: $u = \tan(x)$

$$I = \int_{0}^{\infty} \frac{\ln\left|u^2 + 1 + u^4 \right|}{u^2 + 1}\:du = \int_{0}^{\infty} \frac{\ln\left|1 + u^2\left(u^2 + 1\right) \right|}{u^2 + 1}\:du$$

(2) Now employ Feynman's Trick by introducing a new parameter:

$$I(t) = \int_{0}^{\infty} \frac{\ln\left|1 + t^2u^2\left(u^2 + 1\right) \right|}{u^2 + 1}\:du$$

Note here that $I = I(1)$ and $I(0) = 0$

(3) Take the derivative w.r.t 't'

$$I'(t) = \int_{0}^{\infty} \frac{2tu^2\left(u^2 + 1\right)}{1 + t^2u^2\left(u^2 + 1\right)}\frac{1}{u^2 + 1}\:du = \frac{1}{t} \int_{-\infty}^{\infty} \frac{1}{\left(u - \frac{1}{tu}\right)^2 + \frac{2}{ t} + 1}\:du$$

(4) Employ Glasser's Master Theorem:

$$I'(t) = \frac{1}{t} \int_{-\infty}^{\infty} \frac{1}{\left(u - \frac{1}{tu}\right)^2 + \frac{2}{t} + 1} \:du= \frac{1}{t}\int_{-\infty}^{\infty}\frac{1}{u^2 + \frac{2}{t} + 1} \:du$$

As: $\frac{2}{t} + 1 > 0 $ we arrive at

$$I'(t) = \frac{1}{t}\left[\frac{1}{\sqrt{\frac{2}{t} + 1}}\arctan\left(\frac{u}{\frac{2}{t} + 1}\right)\right]_{-\infty}^{\infty}= \frac{\pi}{\sqrt{t\left(t + 2\right)}}$$

(5) We now integrate w.r.t 't'

$$I(t) = \int \frac{\pi}{\sqrt{t\left(t + 2\right)}}\:dt = 2\pi\sinh^{-1}\left(\frac{t}{\sqrt{2}} \right) + C$$

Where $C$ is the constant of integration. As above $I(0) = 0 \rightarrow C = 0$ and so, our final solution is given by:

$$I = I(1) = 2\pi\sinh^{-1}\left(\frac{1}{\sqrt{2}} \right)$$

  • 1
    $\begingroup$ One equivalent integral is $\int_0^{\pi/2}\log(2+\cot^2x)\,dx$, since $\int_0^{\pi/2}\log(\tan^2x)\,dx=0$. I only raise this since you asked about a very similar integral before [here][1]. [1]: math.stackexchange.com/questions/3007234/… $\endgroup$ Commented Dec 5, 2018 at 0:41
  • $\begingroup$ Well there is an antiderivative: wolframalpha.com/input/… $\endgroup$
    – clathratus
    Commented Dec 5, 2018 at 0:45
  • $\begingroup$ @JamesArathoon - No, I appreciate the comment. Yes very similar as you pointed out. I'm in the process of trying to find commonality between integrals as means of deriving a set of integral structures and their respective methods (or prescribed methods). This has been a rapid pace over the past few months, so you will certainly see that as time moves forward and I'm made aware of different approaches that commonly solved integrals from the past are 'revamped'. $\endgroup$
    – user150203
    Commented Dec 5, 2018 at 1:15

3 Answers 3


$$I = \int_{0}^{\frac{\pi}{2}} \ln\left(\sec^2(x) + \tan^4(x) \right)dx=\int_0^\infty \frac{\ln(1+x^2+x^4)}{1+x^2}dx$$Consider: $$I(a)=\int_0^\infty \frac{\ln((1+x^2)a+x^4)}{1+x^2}dx$$ Derivating under the integral sign with respect to $a$ gives: $$I'(a)=\int_0^\infty \frac{1+x^2}{(1+x^2)a+x^4}\frac{dx}{1+x^2}=\int_0^\infty \frac{1}{x^4+ax^2+a}dx\overset{\large{x=\frac{\sqrt a}{t}}}=\int_0^\infty \frac{\frac{t^2}{\sqrt a}}{t^4+at^2+a}dt$$ $$2I'(a)=\int_0^\infty \frac{\frac{t^2}{\sqrt a}+1}{t^4+at^2+a}dt\Rightarrow I'(a)=\frac{1}{2\sqrt a}\int_0^\infty \frac{t^2+\sqrt a}{t^4+at^2+a}dt$$ $$=\frac{1}{2\sqrt a}\int_0^\infty \frac{1+\frac{\sqrt a}{t^2}}{\left(t-\frac{\sqrt a}{t}\right)^2+a+2\sqrt a}dt=\frac{1}{2\sqrt a}\int_0^\infty \frac{d\left(t-\frac{\sqrt a}{t}\right)}{\left(t-\frac{\sqrt a}{t}\right)^2+\left(\sqrt{a+2\sqrt a}\,\right)^2}$$ $$=\frac{1}{2\sqrt a}\frac{1}{\sqrt{a+2\sqrt a}}\arctan\left(\frac{t-\frac{\sqrt a}{t}}{\sqrt{a+2\sqrt a}}\right)\bigg|_0^\infty \Rightarrow I'(a)=\frac{\pi}{2\sqrt a}\frac{1}{\sqrt{a+2\sqrt a}}$$ And noticing that $I(0)=4\int_0^\infty \frac{\ln x}{1+x^2} dx=0$. By the fundamental theorem of Calculus we have: $$I=I(1)-I(0)=\int_0^1 I'(a)da=\frac{\pi}{2}\int_0^1 \frac{1}{\sqrt a \sqrt {a+2\sqrt a}}da$$ Finally setting $\sqrt a =x$ gives: $$I=\pi \int_0^1 \frac{1}{\sqrt{(x+1)^2-1}}dx=\pi\ln(2+\sqrt 3)$$

  • 1
    $\begingroup$ It seems we have taken similar approaches. Do you happen to know of any links for the integral result that you cite in your result? This could very easily be verified with Glasser's Master Thereom. I wonder though if there are other way to approach the problem that simplify the process even further? $\endgroup$
    – user150203
    Commented Dec 5, 2018 at 0:14
  • 1
    $\begingroup$ I have used the result from here: www-elsa.physik.uni-bonn.de/~dieckman/IntegralsDefinite/… and it's quite similar, although the parameter choice for $I(a)$ is abit different. $\endgroup$
    – Zacky
    Commented Dec 5, 2018 at 0:15
  • $\begingroup$ @DavidG I have edited now with a direct evaluation for $I'(a)$ in order to set things right! Also in your answer seems like $(4)$ is also known as the Cauchy-Schlomilch transformation as seen here: arxiv.org/pdf/1004.2445.pdf $\endgroup$
    – Zacky
    Commented Dec 5, 2018 at 19:25
  • $\begingroup$ ! Cheers mate! much appreciated! $\endgroup$
    – user150203
    Commented Dec 5, 2018 at 23:22

We use the representation

$$ I=\int_0^{\infty}\frac{\log(g(x))}{1+x^2}dx $$

derived by OP.

Here $g(z)=1+z^2+z^4$. Note that $\log(g(z))$ has four branch points at $z_n=e^{i n \pi/3}$, $n={1,2,4,5}$ of which $z_{1,2}$ lie in the upper half of the complex plane. Let us define

$$ f(z)=\frac{\log(g(z))}{1+z^2} $$ By parity we have also that $2\int_0^{\infty}\frac{\log(g(x))}{1+x^2}dx=\int_{-\infty}^{\infty}\frac{\log(g(x))}{1+x^2}dx$. We furthermore note that since $\log(g(z))\sim_i-2i(x-i)$ the residue at $i$ vanishs. Last but not least, $|f(z) |\sim C\log(R)/R^2$ so integrals over large semicirles of this function vanish in the limit of $R\rightarrow \infty$.

We can therefore state that twice our integral of interest equals the two integrals encirceling the two branchcuts in the upper half of the complex plane ($\delta\rightarrow 0_+$).

$$ 2I=\color{blue}{\int_{e^{i \pi(1/3-\delta)}}^{e^{i \pi(1/3-\delta)}\infty}f(z)dz-\int_{e^{i \pi(1/3+\delta)}}^{e^{i \pi(1/3+\delta)}\infty}f(z)dz}-\\ \color{red}{\int_{e^{i \pi(2/3-\delta)}}^{e^{i \pi(2/3-\delta)}\infty}f(z)dz-\int_{e^{i \pi(2/3+\delta)}}^{e^{i \pi(2/3+\delta)}\infty}f(z)dz}$$ It is a well known fact that such pairs of integrals collapse into integrals over the discontinuity of the integrand which is given in both cases by $2 \pi i\times(1+z^2)^{-1}$ and therefore: $$ 2I=2\pi i\left[\color{blue}{z_1\int_1^{\infty}\frac{dq}{1+(z_1q)^2}}-\color{red}{z_2\int_1^{\infty}\frac{dq}{1+(z_2q)^2}}\right]=\\ 2\pi i[\color{blue}{\text{arccot}(z_1)}-\color{red}{\text{arccot}(z_2)}] $$

Annoying algebra yields ($\text{arccot}(z_{1,2})=\mp i\log(2+\sqrt{3})+\frac{\pi}{4}$) the pleasantly simple end result:

$$ I=\pi(\log(2+\sqrt{3})) $$

  • $\begingroup$ Very nicely done. I really need to revise my contour integration stuff. It seems to simplify the process remarkable. $\endgroup$
    – user150203
    Commented Dec 5, 2018 at 1:56
  • $\begingroup$ you are welcome. note also how this can easily generalized to $g(x)=\sum_{n\geq k \geq0} x^{2k}$ and comperable functions $\endgroup$ Commented Dec 5, 2018 at 1:59
  • $\begingroup$ I did a degree in Applied Maths and whilst we touched on it, it was never a massive thing. Now that I'm very much into more 'Pure' Mathematics I've got a lot of catching up to do! If you know of any good resources on contour integration, please let me know. $\endgroup$
    – user150203
    Commented Dec 5, 2018 at 23:21

Here's an alternative solution. First, notice that if $I(a,b) = \int_0^\pi \ln\left(a + b\cos(x)\right) \mathrm{d}x$ then $I(a,0) = \pi \ln(a)$. Thus \begin{align} \frac{\partial I}{\partial a} &=\int_0^\pi \frac{\mathrm{d}x}{a + b\cos(x)} \overset{t=\tan\left(\frac{x}{2}\right)}{=}\frac{2}{a-b}\int_0^\infty \frac{\mathrm{d}t}{t^2 + \frac{a+b}{a-b}} \overset{t =\sqrt{\frac{a+b}{a-b}}\tan(u)}{=} \frac{\pi }{\sqrt{a^2-b^2}} \end{align} So integrating the above equation w.r.t. $a$ we get $$ I(a,b) = \pi\int\frac{\mathrm{d}a}{\sqrt{a^2-b^2}} = \pi\int\frac{1 + \frac{a}{\sqrt{a^2-b^2}}}{a+\sqrt{a^2-b^2}} \,\mathrm{d}a \overset{u =a+\sqrt{a^2-b^2} }{=} \pi\ln\left(a+\sqrt{a^2-b^2}\right) + C $$ Recalling our initial condition, by plugging in $b=0$ into the previous equation we get $\pi \ln(a) = I(a,0) = \pi \ln(2a) + C$. Hence $C = - \pi \ln(2)$ and

$$\int_0^\pi \ln\left(a + b\cos(x)\right) \,\mathrm{d}x = \pi\ln\left(\frac{a+\sqrt{a^2-b^2}}{2}\right) \tag{1}$$

Lastly, using basic integral manipulations we can obtain $\int_0^{\frac{\pi}{2}}\ln(\cos(x)) \mathrm{d}x = - \frac{\pi}{2}\ln(2)$. Combining everything we conclude the problem as follows

\begin{align} \int_0^{\frac{\pi}{2}} \ln\left(\sec^2(x) + \tan^4(x)\right) \mathrm{d}x & =\int_0^{\frac{\pi}{2}} \ln\left(\frac{7+\cos(4x)}{8\cos^4(x)}\right) \mathrm{d}x \\ & = \frac{1}{2}\int_0^{\pi}\ln(7+\cos(u))\, \mathrm{d}u - 4\int_0^{\frac{\pi}{2}} \ln(\cos(x)) \mathrm{d}x - \frac{\pi}{2}\ln(8)\\ & \overset{(1)}{=}\frac{\pi}{2}\ln\left( \frac{7+4 \sqrt{3}}{2}\right) + 2\pi \ln(2) - \frac{3\pi}{2}\ln(2)\\ & = \boxed{\frac{\pi}{2} \ln\left(7 + 4 \sqrt{3}\right)} \end{align} which agrees with the other answers since $\sqrt{7 + 4 \sqrt{3}} = 2 + \sqrt{3}$.


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