# Problem involving Fundamental Theorem of Calculus: simplifying $\frac{\mathrm d}{\mathrm dx}\int_x^{x^2}\frac{t}{\log t}\,\mathrm dt$

I'm working on the following question:

Simplify the following: $$\frac{\mathrm d}{\mathrm dx}\int_x^{x^2}\frac{t}{\log t}\,\mathrm dt$$

The solution key says this simplifies to: $$2x\frac{x^2}{\log(x^2)}-\frac{x}{\log(x)}$$

I think this wrong though. For two reasons:

1. We can't just use fundamental theorem of calculus (FTC) because the integrand needs to be continuous on the interval of interest. The quotient of continuous functions is continuous, provided the denominator is non-zero. There's a problem though at $$t=1$$.
2. FTC is stated with one end fixed -- can we just assume both ends are functions?

Given these caveats, I'm not sure how to proceed -- thoughts?

• They don't write in the question what should be the domain of the function?
– Mark
Commented Aug 16, 2018 at 16:08
• No more information is given! It's from a crappy math gre test prep book. I guess I should also ask: how should this question be asked so that the given answer is correct, lol Commented Aug 16, 2018 at 16:12
• It is good to pay attention to these details. Commented Aug 16, 2018 at 22:11

The domain of the function $f(x) = \displaystyle \int_x^{x^2} \dfrac{t}{\log t} \, dt$ is $(0,1) \cup (1,\infty)$, but it would certainly be consistent to define $f(1) = 0$ since that integral is defined on a degenerate interval. Take the domain to be $(0,1) \cup (1,\infty)$.

Now, let $F$ be an antiderivative of $\dfrac{t}{\log t}$ defined on either interval $(0,1)$ or $(1,\infty)$. The fundamental theorem of calculus gives you $$f(x) = F(x^2) - F(x)$$ so that $$f'(x) = F'(x^2) \cdot 2x - F'(x) = \frac{x^2}{\log (x^2)} \cdot 2x - \frac{x}{\log x}.$$

1. It is probably implicitly assumed that $x \in (0,\infty)\backslash \{1\}$. (Note that this guarantees that both $x$ and $x^2$ are either in $(0,1)$ or in $(1,\infty)$.)

2. You can split the integral into two parts, i.e. you can write $$\int_x^{x^2} \frac{t}{\log t} \, dt = \int_x^{a} \frac{t}{\log t} \, dt + \int_a^{x^2} \frac{t}{\log t} \, dt = \int_a^{x^2} \frac{t}{\log t} - \int_a^{x} \frac{t}{\log t} ,$$ where $a$ is a constant between $x$ and $x^2$, and then use the FTC on each term separately.

• Do we even need the "between" condition? Isn't it sufficient for $a$ to be in the same connected part of the domain as $x,$ that is, if $0<x<1$ then $0<a<1$? Commented Aug 16, 2018 at 20:41
• @DavidK You're definitely right. But then I would need to specify each case separately. Simply saying that $a$ is between $x$ and $x^2$ allowed me to treat them both simultaneously.
– MSDG
Commented Aug 16, 2018 at 20:54
• I don’t think any new cases are needed. The integral $\int_a^b$ where $a>b$ has a well-known definition, and the integral can be written as a sum of two integrals in exactly the same way. Commented Aug 16, 2018 at 22:23

Notice that the integrand is only defined for $\mathbb R^+\backslash\{1\}$. Therefore the region $(x, x^2)$ never crosses $1$.

To your second question, we can always split the integral anywhere we like. For example, when $x \in (3,9)$,

$$\int^{x^2}_x \frac{t}{\log t} \mathrm dt=\int^{x^2}_9 \frac{t}{\log t} \mathrm dt+\int^{9}_x \frac{t}{\log t} \mathrm dt$$

... and you can proceed by differentiating them seperately. When $x$ is not in this region, you can always choose some other splitting point, and sew the results together.