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I was unable to find an explicit expression for an antiderivative involving an inverse function: Let $h(z)$ be a function (from $\mathbb{R}_+\to\mathbb{R}_+$, monotonously increasing) with antiderivative $H(z)$. The integral I'm looking for is $$\int \frac{h^{-1}(z)}{z} \,\text{d}z\,.$$ I didn't manage to adapt the formula for the simpler $\int h^{-1}(z)\,\text{d}z$ (see e.g. here) to this situation, and integration by parts didn't lead me anywhere either.

Is there a general formula for this integral (possibly with some assumptions) in terms of $h$, $h^{-1}$ and $H$?

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Toffomat
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You can set $t=h^{-1}(z)$ and write $$\int_a^b \frac{h^{-1}(z)}{z}\,dz=\int_{h^{-1}(a)}^{h^{-1}(b)}\frac{t\cdot h'(t)}{h(t)}\,dt=\left[t\ln h(t)\right]_{h^{-1}(a)}^{h^{-1}(b)}-\int_{h^{-1}(a)}^{h^{-1}(b)}\ln h(t)\,dt=\\ =\left[h^{-1}(t)\ln t\right]_a^b-\int_{h^{-1}(a)}^{h^{-1}(b)}{\ln h(t)\,dt}=\left[h^{-1}(t)\ln t\right]_a^b-\left[V_h(h^{-1}(t)\right]_a^b$$

Where $V_h(t)=\int \ln h(z)\,dz$. If you had a fomula in terms of $H, h, h^{-1}$, elementary functions and composition of functions for your integral, then you'd have a formula of the same kind for $V_h(t)$. But, in general, you do not. For instance, $\int \ln\ln (x+3)=(x+3)\ln\ln(x+3)-\operatorname{Li}(x+3)$ and, well... there appears not to be much to work around $\operatorname{Li}$.

  • Ah, that's what I feared. It's always interesting to see how small changes in the problem can strongly change the result... – Toffomat Oct 20 '16 at 07:20