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  • Evaluating   <img alt="\mathrm{\lim _{x \rightarrow \infty} x \int_0^{\pi / 4} \ln \left(1+\tan ^x t\right) d t}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%2

Evaluating   \mathrm{\lim _{x \rightarrow \infty} x \int_0^{\pi / 4} \ln \left(1+\tan ^x t\right) d t}

Option: 1

\mathrm{\frac{\pi^2}{24}}


Option: 2

\mathrm{\frac{\pi^2}{2}}


Option: 3

0


Option: 4

None of these


Answers (1)

best_answer

Applying the substitution  \mathrm{u=\tan ^x t}, we find that

                      \mathrm{x \int_0^{\frac{x}{4}} \log \left(1+\tan ^x t\right) d t=\int_0^1 \frac{\log (1+u)}{u} \cdot \frac{u^{1 / x}}{1+u^{2 / x}} d u}.

By the dominated convergence theorem, as \mathrm{x\rightarrow \infty} we have

                     \mathrm{\lim _{x \rightarrow \infty} x \int_0^{\frac{x}{4}} \log \left(1+\tan ^x t\right) d t=\frac{1}{2} \int_0^1 \frac{\log (1+u)}{u} d u=\frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{24}}.

Posted by

Deependra Verma

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