Integration by substitution, $\int_1^a f(x^s) dx = \int_1^{a^s} f(x) \dfrac{1}{sx^{1-1/s}} dx$

Question

I am trying to show that, for a function $f: \mathbb{R} \to \mathbb{R}$, and any $s>0$, $a>1$, we have that $$ \int_1^a f(x^s) dx = \int_1^{a^s} f(x) \dfrac{1}{sx^{1-1/s}} dx.$$

I am trying to use the following result (integration by substitution theorem): $$\int_c^d f(t) dt = \int_{\phi^{-1}(c)}^{\phi^{-1}(d)}f(\phi(t)) \phi'(t) dt,$$ where $\phi$ is a continuously differentiable bijective function.

My attempt is failing as follows: Let $\phi(t)=t^{1/s}$. Then $\phi(1)=1$ and $\phi(a^s)=a$. Also, $\phi'(t)=\dfrac{1}{s}t^{1/s-1}.$ At this point, I go awry: $$ \int_1^a f(x^s) dx = \int_1^{a^s} f(\phi(x^s)) \phi'(x^s) dx = \int_1^{a^s}f(x)\dfrac{1}{s}(x^s)^{1/s-1} dx.$$ This is incorrect. I have tried to rectify it by using: $\phi(x^s)=x$ and so $\phi'(x^s) \times sx^{s-1}=1$.

However, this gives $$\int_1^{a^s} f(x) \dfrac{1}{sx^{s-1}}dx.$$

Both attempts I have made are incorrect. Can someone help me understand how I am going wrong here? I am grateful for your help. Thank you.

Answer 1

Note the following: In your case $\phi(t) = t^s$. So write: $$ \int^a_1 f(x^s)~\mathrm{d}x = \int^a_1 f(\phi(t))~\mathrm{d}t = \int^a_1 f(\phi(t)) \frac{\phi'(t)}{\phi'(t)}~\mathrm{d}t = \int^a_1 g(\phi(t))\phi'(t)~\mathrm{d}t $$ We had set $g(\phi(t)) := \frac{f(\phi(t))}{\phi'(t)}$. This seems a little weird at first, but we will just accept it as of now. NOW use substitution $u = t^s$: $$ \int^a_1 g(\phi(t))\phi'(t)~\mathrm{d}t = \int^{a^s}_1 g(u)~\mathrm{d}u $$ The difficulty is to write $\frac{f(\phi(t))}{\phi'(t)}$ in terms of $u$. Take a look: $$ \frac{f(\phi(t))}{\phi'(t)} = \frac{f(t^s)}{st^{s-1}} = \frac{f(u)}{st^{s-1}} $$ Since $u=t^s$, we get $t = \sqrt[s]{u}$. So: $$ \frac{f(u)}{st^{s-1}} = \frac{f(u)}{su^{1-\frac{1}{s}}} $$ This means: $$ \int^a_1 f(x^s)~\mathrm{d}x = \int^{a^s}_1 \frac{f(u)}{su^{1-\frac{1}{s}}}~\mathrm{d}u $$

This way - and only this way - is how substitution works. The delicate part is the reciprocal derivative that usually (also in this case) appears. You have to artificially add it in order to be able to use the formula.

Your choice of $u$ was wrong in the first place. And if you call a variable $u$, you should stick to its name all the way.