definite integral

[man111]

If $\alpha,\beta,\gamma,\delta >0$

Then prove that $\displaystyle \int_{e^{\gamma}}^{e^{\delta}}\left(f(x^{\beta})-f(x^{\alpha})\right)\frac{dx}{x\ln (x)}=\int_{e^{\alpha}}^{e^{\beta}}\left(f(x^{\delta})-f(x^{\gamma})\right)\frac{dx}{x\ln (x)}$

[Dave]

Let $h(x)= \int \frac {f(e^x)} x \, dx$ - antiderivative of $\frac {f(e^x)} x$. Then $\frac {dh} {dx} = \frac {f(e^x)} x$ and, for any $\theta > 0, \, x > 0$: $\frac d {dx} \, h(\theta \ln x) = \frac {f(e^{\theta \ln x})} {\theta \ln x} \frac {\theta} x = \frac {f(x^{\theta})} {x \ln x}$. So $\int \frac {f(x^{\theta})} {x \ln x} \, dx = h(\theta \ln x)$ and

$$\int_{e^{\gamma}}^{e^{\delta}}\left(f(x^{\beta})-f(x^{\alpha})\right)\frac{dx}{x\ln (x)} = \int_{e^{\gamma}}^{e^{\delta}} f(x^{\beta}) \frac{dx}{x\ln (x)} \; - \; \int_{e^{\gamma}}^{e^{\delta}} f(x^{\alpha}) \frac{dx}{x\ln (x)} = h(\beta \delta) - h(\beta \gamma) - h(\alpha \delta) + h(\alpha \gamma).$$

The last expession is symmetric over substitution $(\alpha, \beta) \leftrightarrow (\gamma, \delta)$.

[ewert]

$$F(\alpha,\beta)\equiv\int\limits_{\gamma}^{\delta}\big(f(e^{\alpha t})-f(e^{\alpha t})\big)\,\dfrac{dt}t=\Big[?\Big]=\int\limits_{\alpha}^{\beta}\big(f(e^{\gamma t})-f(e^{\delta t})\big)\,\dfrac{dt}t\equiv G(\alpha,\beta);$$

$$\dfrac{\partial f(e^{\alpha t})}{\partial\alpha}=\dfrac{t}{\alpha }\cdot\dfrac{\partial f(e^{\alpha t})}{\partial t};$$

$$\dfrac{\partial F(\alpha,\beta)}{\partial\alpha}=\dfrac1{\alpha}\int\limits_{\gamma}^{\delta}\dfrac{\partial f(e^{\alpha t})}{\partial t}\,dt=\dfrac1{\alpha}\big(f(e^{\alpha\delta})-f(e^{\alpha\gamma})\big)\equiv\dfrac{\partial G(\alpha,\beta)}{\partial\alpha}.$$

Этого достаточно, т.к. $F(\beta,\beta)\equiv G(\beta,\beta)\equiv0.$

[man111]

thank dave, ewert