# Finding a function $g$ satisfying $\int_0^{x^2} \operatorname{tg}(t)dt=x+x^2$

I want to find function(s) $$g$$ satisfying $$\int_0^{x^2} \operatorname{tg}(t)dt=x+x^2 \tag{1}$$

*Assuming that $$\operatorname{tg}(t)$$ is continuous on $$[0,x^2]$$, fundamental theorem of calculus gives: $$2x^3g(x^2)=1+2x \\\Rightarrow g(x^2)= 0.5 x^{-3}+x^{-2} \;\text{for all x\ne0}\;\\\Rightarrow g(t)=0.5t^{-3/2}+t^{-1} \;\;\forall t\gt 0$$ $$\tag{2}$$

Defining $$g$$ as follows: $$g=\begin{cases} 0.5t^{-3/2}+t^{-1} & t\gt 0 \\[12pt] 1 & t\le0 \end{cases} \\\Rightarrow \operatorname{tg}(t)=\begin{cases} 0.5t^{-1/2}+t & t\gt 0 \\[12pt] t & t\le0 \end{cases}$$

For all $$x,\operatorname{tg}(t)$$ satisfies $$(1)$$. I am unable to understand the following:

Question 1: Why this $$g(t)$$ is an acceptable solution because clearly $$\operatorname{tg}(t)$$ is not continuous at $$0$$ hence violating our assumption *, in which case we can not apply Fundamental Theorem of Calculus because the integrand is not continuous on $$[0,x]$$.

Question 2: How can it be shown/proven that $$g(t)$$ above is the only solution.

Can anyone please help me on this? Any help on this will be highly appreciated. Thanks in advance.

• Upper bound $x$ or $x^2$ ?? – Yves Daoust Apr 13 '20 at 20:18
• @YvesDaoust, Thanks for pointing that out. I have edited my post. – Koro Apr 13 '20 at 20:20

Although $$t g(t)$$ is not continuous on $$[0,1]$$, the improper integral in (1) still exists. Moreover, for any $$c$$ with $$0 < c \le x^2$$, $$\int_0^{x^2} t g(t)\; dt = \int_0^c t g(t)\; dt + \int_c^{x^2} t g(t)\; dt$$, and the integrand is continuous on $$[c, x^2]$$, so you can apply the FTC there.