From $X:S^1 \to S^2$ you can construct the loop $X' : S^1 \to S^2 \setminus \{pt\}$ because $X$ misses $pt$. Then, since you have $g : S^2 \setminus \{pt\} \to \Bbb R^2$ and $h : \Bbb R^2 \to S^2 \setminus \{pt\}$ being inverses of each other, you obtain the loop $X'' = g \circ X': S^1 \to \Bbb R^2$. Since $\Bbb R^2$ is simply connected, you obtain a homotopy $X''' : S^1 \times I \to \Bbb R^2$. Let $\iota : S^2 \setminus \{pt\} \to S^2$ be the inclusion map.
Then, $X'''' : S^1 \times I \to S^2$ gives you the homotopy of $X$ to a constant map, where $X'''' = \iota \circ h \circ X'''$

Ok @TedShifrin, here you go!
We have
$$\int_{-\infty}^{\infty}\frac{ue^{u}e^{yu}}{(1+e^{2u})(1+e^{yu})}du$$
$$ =\int_{-\infty}^{0}\frac{ue^{u}e^{yu}}{(1+e^{2u})(1+e^{yu})}du + \int_{0}^{\infty}\frac{ue^{u}e^{yu}}{(1+e^{2u})(1+e^{yu})}du$$

Looking at the first integral, substitute $u = -v$. We get $du = -dv$. The integral becomes
$$-\int_{\infty}^{0}\frac{(-v)e^{-v}e^{-yv}}{(1+e^{-2v})(1+e^{-yv})}dv$$
$$=-\int_{0}^{\infty}\frac{v}{(e^{v} + e^{-v})(1+e^{yv})(1+e^{-yv})}dv$$
Multiplying through by $\frac{e^{v}e^{yv}}{e^{v}e^{yv}}$ gives us

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