Question: Let $u(t),v(t)$ be smooth curves in a vector space $V$ with inner product $\langle,\rangle$. Let $\dot{v}(t)$ be the differential of the curve at *t*, where we consider $V$ as a manifold. Since for a vector space manifold we can identify each tangent space with $V$ itself, $\dot{v}(t)$ is a curve in $V$.

I'm reading a paper where they seem to use a formula similar to

$\int\_{0}^{T} \langle u(t), \dot{v}(t) \rangle dt = - \int\_{0}^{T} \langle \dot{u}(t), v(t) \rangle dt + something$.

Could someone point me to the exact formula, and its proof?

Thank you.

Answer: 
Substitute $u = x^4$ to get

$$ \int\_{0}^{1} \frac{1-\sqrt{1-x^4}}{x^2\sqrt{1-x^4}} \, \mathrm{d}x
= \frac{1}{4} \int\_{0}^{1} \frac{1-(1-u)^{1/2}}{u^{5/4}(1-u)^{1/2}} \, \mathrm{d}u. $$

Now we regularize the integral by replacing $u^{5/4}$ by $u^s$ for $s < 1$. Then

\begin{align\*}
\frac{1}{4} \int\_{0}^{1} \frac{1-(1-u)^{1/2}}{u^{s}(1-u)^{1/2}} \, \mathrm{d}u
&= \frac{1}{4}\left(B\left(1-s,\tfrac{1}{2}\right)-\frac{1}{1-s}\right) \\
&= \frac{1}{4}\left(\frac{\Gamma(1-s)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{2}-s\right)}-\frac{1}{1-s}\right).
\end{align\*}

Note that the left-hand side defines an analytic function for $\operatorname{Re}(s) < 2$, while the right-hand side is a meromorphic function on all of $\mathbb{C}$ with the removable singularity at $s = 1$. So both sides give rise to analytic functions on $\operatorname{Re}(s) < 2$, and so, they are equal on this region by the principle of analytic continuation. Therefore we can simply plug $s = 5/4$ to get

\begin{align\*}
\frac{1}{4} \int\_{0}^{1} \frac{1-(1-u)^{1/2}}{u^{5/4}(1-u)^{1/2}} \, \mathrm{d}u
&= 1 + \frac{\Gamma\left(-\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{4\Gamma\left(\frac{1}{4}\right)} \\
&= 1 - \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{4}\right)},
\end{align\*}

which simplifies to the desired expression by Euler's reflection formula $\Gamma(z)\Gamma(1-z) = \pi\csc(\pi z)$.
I begin by making the following observation:
$$\frac{d}{dx} \left (\frac{1 - \sqrt{1 - x^4}}{x} \right )= \frac{x^4 + 1 - \sqrt{1 - x^4}}{x^2 \sqrt{1 - x^4}}.$$

So, for your integral, one can write
\begin{align}
\int\_0^1 \frac{1 - \sqrt{1 - x^4}}{x^2 \sqrt{1 - x^4}} \, dx &= \int\_0^1 \frac{(x^4 + 1 - \sqrt{1 - x^4}) - x^4}{x^2 \sqrt{1 - x^4}} \, dx\\
&= \int\_0^1 \left (\frac{1 - \sqrt{1 - x^4}}{x} \right )' \, dx -\underbrace{\int\_0^1 \frac{x^2}{\sqrt{1 - x^4}} \, dx}\_{x \, \mapsto \, \sqrt[4]{x}}\\
&= 1 - \frac{1}{4} \int\_0^1 x^{-\frac{1}{4}} (1 - x)^{-\frac{1}{2}} \, dx\\[1ex]
&= 1 - \frac{1}{4} \operatorname{B} \left (\frac{3}{4}, \frac{1}{2} \right )\\[1ex]
&= 1 - \frac{1}{4} \frac{\Gamma (\frac{3}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{5}{4})}\\
&= 1 - \sqrt{\pi} \frac{\Gamma (\frac{3}{4})}{\Gamma (\frac{1}{4})}\\
&= 1 - \frac{\pi \sqrt{2 \pi}}{\Gamma^2 (\frac{1}{4})},
\end{align}
as required. Here $\Gamma (\frac{3}{4}) \Gamma (\frac{1}{4}) = \pi \sqrt{2}$, has been used.