Prove the identity $\int_{2a-1}^{2a} f(x)dx = 2 \int_{a}^{\infty}g(x)dx$

Question

I'm struggling with an integral. I'm allowed to use Maple, but even that doesn't help.

I have the following functions:

$$f(x)=\frac{x}{1+x^{4/3}}\ln(1+\ln x)$$ and $$g(x)=f(2x-1)-f(2x)$$ And I wish to show: $$\int_{2a-1}^{2a} f(x)dx = 2 \int_{a}^{\infty}g(x)dx$$ I already tried integration by parts for the LHS, but that went wrong. Can somebody help me please?

best wishes, Joe

Answer 1

Assume the integral on the left exists. Then it can be formally written as:

$$\int_{2a-1}^{2a} f(x)dx=F(2a)-F(2a-1)$$

Where:

$$F'(x)=f(x)$$

Define:

$$G'(x)=g(x)$$

Now differentiate:

$$\frac{d}{dx} (F(2x)-F(2x-1))=2 f(2x)-2 f(2x-1)=-2 g(x)$$

So we can write, up to a constant:

$$G(x)=\frac{1}{2} (F(2x-1)-F(2x))$$

You see where I'm going with this?

Further hint: you only need the explicit form for $f(x)$ to deal the limit for $G(x)$ at $x \to \infty$.