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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,

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

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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}

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  • $\begingroup$ Integral of $t^{n-1}$ is $\frac{t^n}{n}$.How did you compute $\frac{t^n-1}{n}$? $\endgroup$ – Dhamnekar Winod Jul 12 '18 at 13:45
  • $\begingroup$ @DhamnekarWinod I just added a constant, so it is still an antiderivative/indefinitie integral. $\endgroup$ – ComplexYetTrivial Jul 12 '18 at 13:50
  • $\begingroup$ ,When I integrated$n\int_0^1t^{n-1}ln(1-t)$ i got $t^n-1ln(1-t)-\int_0^1\frac{t^n-1}{1-t}$. How did you compute$-\int_0^1\frac{1-t^n}{1-t}$ $\endgroup$ – Dhamnekar Winod Jul 12 '18 at 14:44
  • $\begingroup$ If you substitute the limits $0$ and $1$ into the first term, $(t^n - 1) \ln (1-t)$, you will find that it vanishes. And in the second term you should get an additional minus sign from the chain rule: $ \frac{\mathrm{d}}{\mathrm{d} t} \ln(1-t) = \color{red}{-} \frac{1}{1-t}$ . $\endgroup$ – ComplexYetTrivial Jul 12 '18 at 15:04
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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 $0$.

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  • $\begingroup$ Shoud it be: $\int\frac{(1-u)^n}{u}\,du= \ln u+\sum_{k=1}^n(-1)^k\binom{n}{k}\frac{u^{k+1}}{k(k+1)}$? $\endgroup$ – Oldboy Jul 12 '18 at 9:07
  • $\begingroup$ @Oldboy You forgot to fully divide by $u$. $\endgroup$ – egreg Jul 12 '18 at 9:08
  • $\begingroup$ @egreg Sorry, you're right :) $\endgroup$ – Oldboy Jul 12 '18 at 9:09
  • $\begingroup$ @ClaudeLeibovici Wolframalpha says the summation is $-\psi(n+1)-\gamma$, with $\psi$ the digamma function and $\gamma$ the Euler-Mascheroni constant. $\endgroup$ – egreg Jul 12 '18 at 9:12
  • $\begingroup$ @egreg my $\frac{du}{dx}=-\frac{1}{cos^2x*-n*((tanx)^{\frac{n-1}{n}})}$ But your$\frac{du}{dx}$ is different.How is that? $\endgroup$ – Dhamnekar Winod Jul 12 '18 at 14:54

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