I am trying to evaluate this integral

$$I=\large\int_{0}^{\pi}\sec(x)\sqrt{\tan\left(\frac{x}{2}\right)}\ln^n\tan\left(\frac{x}{2}\right)\mathrm dx$$

Making a substitution $\large u=\frac{x}{2}$

$\large \mathrm dx=\mathrm 2du$

$$I=2\large \int_{0}^{\pi/2}\sec(2u)\sqrt{\tan(u)}\ln^n\tan(u)\mathrm du$$

Using trigonometrical identities, making a transformation

$$I=-2\large \int_{0}^{\pi/2}\sec(2u)\cdot \frac{\sqrt{\tan(u)}\ln^n\tan(u)}{\tan^2(u)-1}\mathrm du$$

Doing another substitution $\large v=\tan(u)$

$\large \mathrm du=\frac{1}{\sec^2(u)}\mathrm dv$

$$\large I=-2\int_{0}^{\infty}\frac{\sqrt{v}\ln^n(v)}{v^2-1}\mathrm dv$$

Doing another substitution $\large y=\sqrt{v}$

$\large \mathrm dv=2\sqrt{v}\mathrm dy$

$$\large I=-64\int_{0}^{\infty}\frac{y^2\ln^n(y)}{y^4-1}\mathrm dy$$

$$\large I=-64\int_{0}^{\infty}\frac{y^2\ln^n(y)}{(y^2+1)(y-1)(y+1)}\mathrm dy$$

Using partial fraction decomposition

$$\large I=-32\int_{0}^{\infty}\frac{\ln^n(y)}{y^2+1}\mathrm dy+16\int_{0}^{\infty}\frac{\ln^n(y)}{y+1}\mathrm dy-16\int_{0}^{\infty}\frac{\ln^n(y)}{y-1}\mathrm dy$$

I can't continue from this point... I would like some help, please.

  • 2
    $\begingroup$ Where did you get this horrible integral from? $\endgroup$ – TheSimpliFire Jul 15 '18 at 9:28
  • 1
    $\begingroup$ I have worked through this with $n=1$ and got $-\pi^2$ but can't get anywhere with any other power! Where did you find this integral?? $\endgroup$ – Mandelbrot Jul 15 '18 at 10:09
  • $\begingroup$ Use residue theorem to get recursive relation. $\endgroup$ – Szeto Jul 15 '18 at 10:43
  • $\begingroup$ writing your integral as a sum of divergent integrals may not be a good strategy. $\endgroup$ – GEdgar Jul 15 '18 at 12:58

Split the integral at $v=1$ and let $v \to \frac{1}{v}$ in the second integral to obtain $$ I_n \equiv - 2 \int \limits_0^\infty \frac{\sqrt{v} \ln^n (v)}{v^2 - 1} \, \mathrm{d} v = -2 \int \limits_0^1 \frac{[- \ln(v)]^n}{1-v^2} \left[v^{-1/2} + (-1)^{n-1} v^{1/2}\right] \, \mathrm{d} v \, . $$ Now let $v = \mathrm{e}^{-t}$ and expand the denominator into a geometric series: $$ I_n = -2 \int \limits_0^\infty t^n \sum \limits_{l=0}^\infty \mathrm{e}^{-(2l+1)t} \left[\mathrm{e}^{t/2} + (-1)^{n-1} \mathrm{e}^{-t/2}\right] \, \mathrm{d} t \, . $$ Interchange summation and integration (monotone convergence theorem), make linear substitutions in the exponents and use the definition of the gamma function to find $$ I_n = -2 n! \sum \limits_{l=0}^\infty \left[\frac{1}{\left(2l + \frac{1}{2}\right)^{n+1}} + \frac{(-1)^{n-1}}{\left(2l + \frac{3}{2}\right)^{n+1}}\right] = - 2^{n+2} n! \sum \limits_{m=0}^\infty \frac{(-1)^{(n-1) m}}{(2m+1)^{n+1}} \, .$$ For odd values of $n$ the sum is given by the Dirichlet lambda function: \begin{align} I_{2k-1} &= -2^{2k+1} (2k-1)! \lambda (2k) = -2^{2k+1} (2k-1)! \left(1-2^{-2k}\right) \zeta (2k) \\ &= - \frac{4^k (4^k-1) \lvert \mathrm{B}_{2k} \rvert}{2 k} \pi^{2k} \, , \, k \in \mathbb{N} \, . \end{align} $\mathrm{B}_{2k}$ are the Bernoulli numbers. For even $n$ one can use the Dirichlet beta function: $$I_{2k} = - 2^{2k+2} (2k)! \beta(2k+1) = - \lvert \mathrm{E}_{2k} \rvert \pi^{2k+1} \, , \, k \in \mathbb{N} \, . $$ $\mathrm{E}_{2k}$ are the Euler numbers. In general, your integral is given by $$ I_n = - a_n \pi^{n+1} \, , \, n \in \mathbb{N} \, ,$$ where $(a_n)_{n \in \mathbb{N}_0}$ is this sequence.

Note that these results suggest $I_0 = - \pi$, which is indeed true if we take the Cauchy principal value: $$ I_0 \equiv - 2 \operatorname{PV} \int \limits_0^\infty \frac{\sqrt{v}}{v^2 - 1} \, \mathrm{d} v = -2 \int \limits_0^1 \frac{\mathrm{d}v}{\sqrt{v} (1+v)} = - 4 \int \limits_0^1 \frac{\mathrm{d}w}{1+w^2} = - \pi \, . $$

  • $\begingroup$ (+1) for the great work. You and this user math.stackexchange.com/users/9340/sangchul-lee (@Sangchu lee), are magicians in integrals. Both of you guys like integrals:-wrote on the profiles. $\endgroup$ – user565198 Jul 15 '18 at 11:23
  • $\begingroup$ @Bonjour Thank you! Though compared to him and many other users who have been solving crazy integrals on this site for years, I'd say I'm more like a kid that has just found a few cool card tricks on the internet ;) $\endgroup$ – ComplexYetTrivial Jul 15 '18 at 11:47
  • $\begingroup$ How did you get $$-2n!\sum_{l\geq0}\bigg[\frac1{(2l+\frac12)^{n+1}}+\frac{(-1)^{n-1}}{(2l+\frac32)^{n+1}}\bigg]=-2^{n+2}n!\sum_{m\geq0}\frac{(-1)^{m(n-1)}}{(2m+1)^{n+1}}$$ $\endgroup$ – clathratus Nov 12 '18 at 23:03
  • 1
    $\begingroup$ @clathratus Extract the factor $(1/2)^{n+1}$ from the denominator. Then the remaining terms are of the form $1/(4l+1)^{n+1}$ (with a positive sign) and $1/(4l+3)^{n+1}$ (with sign $(-1)^{n-1}$). Thus all odd integers appear in the denominator and the $(-1)^{m(n-1)}$ in the numerator ensures that they have the correct sign. $\endgroup$ – ComplexYetTrivial Nov 13 '18 at 10:03

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