Compute $\int_{0}^{\pi/4}\ln(1-\sqrt[n]{\tan x})\frac{dx}{\cos^2(x)}$

Question

I am trying to compute this

$$ \int_{0}^{\pi/4}\ln(1-\sqrt[n]{\tan x})\frac{\mathrm dx}{\cos^2(x)},\qquad (n\ge1). $$

Making a transformation of $I$ to utilise a sub of $u=1-\sqrt[n]{\tan x}$

\begin{align} I&=\int_{0}^{\pi/4}\frac{\sec^2(x)}{n\sqrt[n]{\tan x}}\cdot n\sqrt[n]{\tan x}\cdot\ln(1-\sqrt[n]{\tan x})\,\mathrm dx \\[6px] &\qquad\mathrm dx=-\frac{n\sqrt[n]{\tan x}}{\sec^2(x)}\,\mathrm du \\[6px] I&=n\int_{0}^{1}(1-u)^{n-1}\ln u \,\mathrm du \end{align}

This can be easily done by integration by parts, but I seem to shruggle in somewhere in evaluating it,

$$ \int(1-u)^{n-1}\ln u \,\mathrm du= n(1-u)^n\ln u-\frac{1}{n^2}\int \frac{(1-u)^n}{u}\,\mathrm du $$

Answer 1

Your integral is given by the negative of the $n$-th harmonic number: $$ I_n \equiv n \int \limits_0^1 (1-u)^{n-1} \ln (u) \, \mathrm{d} u = - H_n = - \sum \limits_{k=1}^n \frac{1}{k} \, . $$ You can use the substitution $u = 1-t$ and then have a look at this question for a derivation. Here's an alternative route: Use the antiderivative $\frac{t^n - 1}{n}$ of $t^{n-1}$ to integrate by parts directly: \begin{align} I_n &= n \int \limits_0^1 t^{n-1} \ln (1-t) \, \mathrm{d} t = - \int \limits_0^1 \frac{1-t^n}{1-t} \, \mathrm{d} t = - \sum \limits_{k=0}^{n-1} \int \limits_0^1 t^k \, \mathrm{d} t = -\sum \limits_{k=0}^{n-1} \frac{1}{k+1} = - H_n \, . \end{align}

Answer 2

With $u=1-\sqrt[n]{\tan x}$, we get $$ \tan x=(1-u)^n $$ so $$ \frac{1}{\cos^2x}\,dx=-n(1-u)^{n-1}\,du $$ Thus the integral becomes $$ \int_1^0 -n(1-u)^{n-1}\ln u\,du= \int_0^1 n(1-u)^{n-1}\ln u\,du $$ Integrating by parts: $$ -(1-u)^n\ln u+\int\frac{(1-u)^n}{u}\,du $$ We have $$ \frac{(1-u)^n}{u}=\frac{1}{u}\sum_{k=0}^n(-1)^k\binom{n}{k}u^k $$ so that $$ \int\frac{(1-u)^n}{u}\,du= \ln u+\sum_{k=1}^n(-1)^k\binom{n}{k}\frac{u^k}{k} $$ Thus an antiderivative can be written as $$ (1-(1-u)^n)\ln u+\sum_{k=1}^n(-1)^k\binom{n}{k}\frac{u^k}{k} $$ The value at $1$ is $$ \sum_{k=1}^n(-1)^k\binom{n}{k}\frac{1}{k} $$ The value at $0$ (actually the limit) is $0$.

Don't try and evaluate prematurely the integration by parts, because the part $-(1-u)^n\ln u$ doesn't converge at $0$.