Need help to evaluate $\int_0^{\infty}\left(\frac{x}{e^{x^2}+e^{-x^2}}\right)^3 d x$.

Question

The question came into my mind that how far we can go with the power $n $ in $\int_0^{\infty}\left(\frac{x}{e^{x^2}+e^{-x^2}}\right)^n \mathrm d x $ after I had found the exact values for $n=1$ and $n=2$, say

$$I_1= \frac{\pi}{8},$$ $$I_2= \frac{\sqrt{\pi}-\sqrt{2 \pi}}{8 \sqrt{2}} \zeta\left(\frac{1}{2}\right) $$ Proof: $$ \begin{aligned} I_1= & \int_0^{\infty} \frac{x}{e^{x^2}+e^{-x^2}} \textrm{ d} x \\ = & \frac{1}{2} \int_0^{\infty} \frac{1}{e^x+e^{-x}}\textrm{ d}x, \text { where } x^2 \mapsto x \\ =&\frac{1}{4} \int_0^{\infty}\frac{ 1}{\operatorname{cosh }x} \textrm{ d} x \\ =&\frac{1}{4} \int_0^{\infty} \frac{\operatorname{\cosh} x}{1+\sinh ^2 x} \textrm{ d} x \\ =&\frac{1}{4}\left[\tan ^{-1}(\sinh x)\right]_0^{\infty} \\ =&\frac{\pi}{8} \end{aligned} $$

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Using the expansion for $|x|<1$, $$ \frac{1}{(1+x)^2}=\sum_{n=1}^{\infty}(-1)^{n-1} n x^{n-1}, $$ we have $$ \begin{aligned} I_2=& \int_0^{\infty}\left(\frac{x}{e^{x^2}+e^{-x^2}}\right)^2 \textrm{d}x \\ = & \frac{1}{2} \int_0^{\infty} \frac{\sqrt{x}}{\left(e^x+e^{-x}\right)^2} \textrm{d} x, \text { via } x^2 \rightarrow x\\ = & \frac{1}{2} \int_0^{\infty} \frac{e^{-2 x} \sqrt{x}}{\left(1+e^{-2 x}\right)^2} \textrm{d}x \\ = & \frac{1}{2} \sum_{n=1}^{\infty}(-1)^{n-1} n \int_0^{\infty} x^{\frac{1}{2}} e^{-2 n x} \textrm{d}x \end{aligned} $$ Utilising the formula: $\int_0^{\infty} t^{z-1} e^{-s t} d t=\frac{\Gamma(z)}{s^z}$ yields $$ \begin{aligned} I_2 & = \frac{1}{2} \sum_{n=1}^{\infty}\left((-1)^{n-1} n \cdot \frac{\Gamma\left(\frac{3}{2}\right)}{(2 n)^{\frac{3}{2}}}\right)\\ & =\frac{1}{4 \sqrt{2}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{\sqrt n} \\ & =\frac{\pi}{8 \sqrt{2}}\left(\sum_{n=1}^{\infty} \frac{1}{\sqrt n}-2 \sum_{n=1}^\infty \frac{1}{\sqrt{2 n}}\right) \\ & =\frac{\sqrt{\pi}-\sqrt{2 \pi}}{8 \sqrt{2}} \zeta\left(\frac{1}{2}\right) \end{aligned} $$

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Similarly, using the expansion for $|x|<1$ : $\frac{1}{(1+x)^3}=\frac{1}{2} \sum_{n=1}^{\infty}(-1)^{n-1}(n+1) n x^{n-1}$ yields $$ \begin{aligned} I_3 & =\int_0^{\infty}\left(\frac{x}{e^{x^2}+e^{-x^2}}\right)^3 \textrm{d} x \\ & =\frac{1}{2} \int_0^{\infty} \frac{x}{\left(e^x+e^{-x}\right)^3} \textrm{d} x, \text { via } x^2 \rightarrow x\\ & =\frac{1}{2} \int_0^{\infty} \frac{e^{-3 x} x}{\left(1+e^{-2 x}\right)^3} \textrm{d} x \\ & =\frac{1}{2} \sum_{n=1}^{\infty}(-1)^{n-1}(n+1) n \int_0^{\infty} x e^{-(2 n+1) x} \textrm{d} x \\ & =\frac{1}{2} \sum_{n=1}^{\infty}\left((-1)^{n-1}(n+1) n \cdot\frac{\Gamma(2)}{(2 n+1)^2}\right) \\ & =\frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n+1) n}{(2 n+1)^2} \end{aligned} $$ which is divergent and contradictory to the answer $0.0259978$ found in WA.

Could you help me by giving opinions or an alternative? Your help is highly appreciated.

Answer 1

We consider the below integral: $$ I(n) = \int_0^{\infty} \left( \frac{x}{e^{x^2} + e^{-x^2}} \right)^n \, dx $$

For background, recall that the Gamma function is defined by $$ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \, dt, \quad \Re(z) > 0, $$ so that in particular $$ \int_0^\infty t^{z-1} e^{-st} \, dt = \frac{\Gamma(z)}{s^z} $$

Also recall the identity: $$ (1+u)^{-n} = \sum_{m=0}^\infty \binom{n+m-1}{m} (-u)^m $$

The main result is stated below: \begin{align} I(n) = \frac{\Gamma\!\left(\tfrac{n+1}{2}\right)}{2} \sum_{m=0}^\infty (-1)^m \binom{n+m-1}{m} (n+2m)^{-\frac{n+1}{2}} \end{align}

Substitution $t = x^2$. Let $t = x^2$, so that $dx = \frac{dt}{2\sqrt{t}}$, and $x = \sqrt{t}$. Then, \begin{align} I(n) &= \int_0^\infty \frac{x^n}{(e^{x^2} + e^{-x^2})^n} \, dx \\ &= \frac{1}{2} \int_0^\infty \frac{t^{n/2}}{(e^t + e^{-t})^n} t^{-1/2} \, dt \\ &= \frac{1}{2} \int_0^\infty t^{\frac{n-1}{2}} \frac{dt}{(e^t + e^{-t})^n} \end{align}

Hence, $$ I(n) = \frac{1}{2} \int_0^\infty t^{\frac{n-1}{2}} \frac{1}{(e^t + e^{-t})^n} \, dt $$

Rewriting the denominator. Observe that $$ e^t + e^{-t} = e^{-t}(1 + e^{2t}) = e^t(1 + e^{-2t}) $$ We use the form: $$ (e^t + e^{-t})^{-n} = e^{-nt}(1 + e^{2t})^{-n} = e^{nt}(1 + e^{-2t})^{-n} $$

Inserting this into the integral gives: $$ I(n) = \frac{1}{2} \int_0^\infty t^{\frac{n-1}{2}} e^{-nt} (1 + e^{-2t})^{-n} \, dt $$

Now apply the binomial series: $$ (1 + e^{-2t})^{-n} = \sum_{m=0}^\infty \binom{n+m-1}{m} (-e^{-2t})^m = \sum_{m=0}^\infty \binom{n+m-1}{m} (-1)^m e^{-2mt} $$

Substituting this into the integral: $$ I(n) = \frac{1}{2} \int_0^\infty t^{\frac{n-1}{2}} e^{-nt} \sum_{m=0}^\infty \binom{n+m-1}{m} (-1)^m e^{-2mt} \, dt $$

Assuming term-by-term integration is justified: \begin{align} I(n) = \frac{1}{2} \sum_{m=0}^\infty (-1)^m \binom{n+m-1}{m} \int_0^\infty t^{\frac{n-1}{2}} e^{-(n+2m)t} \, dt \end{align}

Using the Gamma function: $$ \int_0^\infty t^{\frac{n-1}{2}} e^{-(n+2m)t} \, dt = \frac{\Gamma\!\left( \frac{n+1}{2} \right)}{(n + 2m)^{\frac{n+1}{2}}} $$

Therefore: $$ I(n) = \frac{1}{2} \sum_{m=0}^\infty (-1)^m \binom{n+m-1}{m} \frac{\Gamma\!\left( \frac{n+1}{2} \right)}{(n + 2m)^{\frac{n+1}{2}}} $$

Finally, factor out the constant: $$ I(n) = \frac{\Gamma\!\left( \frac{n+1}{2} \right)}{2} \sum_{m=0}^\infty (-1)^m \binom{n+m-1}{m} (n + 2m)^{-\frac{n+1}{2}} $$

Numerical verification for the closed form and the integral

\begin{array}{c|c|c} n & \text{closed form } & \text{Integral Expression} \\ \hline 1 & 0.39269908169872415481 & 0.39269908169872415482 \\ 2 & 0.094766002680412 & 0.094766002680412133579 \\ 3 & 0.0259978496360762 & 0.025997849636077179039 \\ 4 & 0.0075402944966602 & 0.007540294496602175356 \\ 5 & 0.002256598131146 & 0.0022565981311454201286 \\ 6 & 0.000689062333003 & 0.00068906233300314074189 \\ 7 & 0.000213374690745 & 0.00021337469074537602378 \\ 8 & 0.0000667573925 & 0.00006675739251729180495 \\ 9 & 0.000021051436954106155279 & 0.000021051436954106155279 \\ 10 & 6.679926256 \times 10^{-6} & 6.6799262559104934239 \times 10^{-6} \\ \end{array}

Answer 2

\begin{align} &\int_0^{\infty}\bigg(\frac{x}{e^{x^2}+e^{-x^2}}\bigg)^3dx \>\>\>\>\>x^2\to x \\ =& \ \frac1{16}\int_0^\infty \frac x{\cosh^3x}dx = \frac1{32}\int_0^\infty \frac x{\sinh x}d(\tanh^2x)\\ \overset{ibp}= &\ \frac1{32}\int_0^\infty \frac x{\cosh x}-\frac{\sinh x}{\cosh^2 x}\ dx= \ \frac1{32} (2G-1) \end{align}

where $\int_0^\infty \frac x{\cosh x}dx=2G$.