$$ f(x) = 1-\int_{-\infty}^{\infty} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2 \pi}} \tanh\left( x-\sqrt{x}y \right)dy. $$

I would like to understand how to get to the solution of the following integral:

$$ g(s)=\frac{1}{2}\int_0^s f(x) dx = s-\int_{-\infty}^{\infty} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2 \pi}} \ln\cosh\left( s-\sqrt{s}y \right)dy $$

I know that

$$ \frac{d}{dx} \ln\cosh\left( x-\sqrt{x}y \right) = (1-\frac{y}{2\sqrt{x}})\tanh(x-\sqrt{x}y)$$

So far I tried the following:

$$ \begin{align} g(s) &= \frac{s}{2} - \frac{1}{2}\int_{-\infty}^{\infty} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2 \pi}} \int_{0}^s (1-\frac{y}{2\sqrt{x}}+\frac{y}{2\sqrt{x}}) \tanh\left( x-\sqrt{x}y \right)dx dy \\ &= \frac{s}{2} -\frac{1}{2}\int_{-\infty}^{\infty} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2 \pi}} \ln\cosh\left( s-\sqrt{s}y \right)dy + \frac{1}{2}\int_{-\infty}^{\infty} \frac{ye^{-\frac{y^2}{2}}}{\sqrt{2 \pi}} \int_{0}^s \frac{\tanh\left( x-\sqrt{x}y \right)}{2\sqrt{x}} dx dy\end{align} $$

But I'm stuck there. The first part equals $1/2$ times the solution. Maybe I should use some substitution in the remaining integral?

  • $\begingroup$ Where did you meet this monster? Looks intimidating $\endgroup$ – Yuriy S Oct 30 '18 at 10:00

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