Evaluating an indefinite integral

Question

I need help evaluating the following indefinite integral explicitly

$$\int \frac{1}{1+t^{2^{-n}}} dt$$

I would appreciate any help

Answer 1

Using the substitutions $t=u^{2^n}$ and $u=\dfrac{v}{1-v}$ and $\mathrm{d}u=\dfrac{\mathrm{d}v}{(1-v)^2}$ , we get $$ \begin{align} \int_0^x\frac1{1+t^{2^{-n}}}\,\mathrm{d}t &=\int_0^{x^{2^{-n}}}\frac1{1+u}\,\mathrm{d}u^{2^n}\\ &=2^n\int_0^{x^{2^{-n}}}\frac{u^{2^n-1}}{1+u}\,\mathrm{d}u\tag{$\ast$}\\ &=2^n\int_0^{x^{2^{-n}}/(1+x^{2^{-n}})}\frac{v^{2^n-1}}{(1-v)^{2^n}}\,\mathrm{d}v\\ &=2^n\,\mathrm{B}\left(\frac{x^{2^{-n}}}{1+x^{2^{-n}}};2^n,1-2^n\right) \end{align} $$ where $\mathrm{B}$ is the Incomplete Beta Function.

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Starting with $(\ast)$, we can verify Ethan's formula: $$ \begin{align} &2^n\int_0^{x^{2^{-n}}}\frac{u^{2^n-1}}{1+u}\,\mathrm{d}u\\ &=-2^n\log\left(1+x^{2^{-n}}\right)+2^n\int_0^{x^{2^{-n}}}\frac{1+u^{2^n-1}}{1+u}\,\mathrm{d}u\\ &=-2^n\left(\log\left(1+x^{2^{-n}}\right)+\sum_{k=1}^{2^n-1}\frac{(-1)^k}{k}x^{k2^{-n}}\right) \end{align} $$

Answer 2

An upper bound is given by ignoring the $1$ in the denominator: $$\int_2^x \frac 1{1+t^m}dt \lt \int_2^x \frac 1{t^m}dt=\left.\frac 1{1-m}\cdot\frac 1{t^{m-1}}\right|_2^x$$ The only time this will not be accurate is when both $m$ and $x$ are small, so much of the area comes where $t^m$ is not too much larger than $1$.

Added: the new version is the same as the original question, with $m=2^{-u}$, though this suggests that $0 \lt m \lt 1$

Answer 3

$\frac{1}{1+t^{\frac{1}{2^n}}} = \frac{1}{t^{\frac{1}{2^n}}}\frac{1}{1+\frac{1}{t^{\frac{1}{2^n}}}} = \frac{1}{t^{\frac{1}{2^n}}} \sum_{k=0}^\infty (-1)^k ( \frac{1}{t^{\frac{1}{2^n}}})^k = \sum_{k=0}^\infty (-1)^k \frac{1}{t^{{\frac{1}{2^n}}(k+1)}}$, and this convergence is uniform for $t \ge 2$.

Hence \begin{eqnarray} \int_2^x \frac{1}{1+t^{\frac{1}{2^n}}} dt &=& \sum_{k=0}^\infty (-1)^k \int_2^x \frac{1}{t^{{\frac{1}{2^n}}(k+1)}} dt \\ &=& \sum_{k=0,\, $k \ne 2^n-1}^\infty (-1)^k \frac{1}{1-{\frac{1}{2^n}}(k+1)} ( \frac{1}{x^{{\frac{1}{2^n}}(k+1)-1}} - \frac{1}{2^{{\frac{1}{2^n}}(k+1)-1}} ) + (-1)^{2^n-1} \ln \frac{x}{2}\\ &=& \sum_{k=0,\, $k \ne 2^n-1}^\infty (-1)^k \frac{1}{1-{\frac{1}{2^n}}(k+1)} ( \frac{x}{x^{{\frac{1}{2^n}}(k+1)}} - \frac{2}{2^{{\frac{1}{2^n}}(k+1)}} ) + (-1)^{2^n-1} \ln \frac{x}{2} \end{eqnarray}

Answer 4

I evaluated it explicitly, but thanks for the help you guys.

$$\int \frac{1}{1+t^{2^{-n}}}dt=-2^n\ln(x^{2^{-n}}+1)+\sum_{k=1}^{2^n-1}\frac{(-1)^{k-1}x^{k2^{-n}}}{k2^{-n}}$$