Fundamental Theorem of Calculus and improper integral (second kind) I have been asked to calculate (if exists) the value of $a$ that makes that the following limit exists and is different to $0$.
$$\lim_{x\to0}\frac{\int_0^{x^2}\frac{\log(1+t)-at}{t}dt}{(1-\cos(x/2))^2}$$
My initial idea was to apply L'Hôpital rule and FTC (and then proceed by equivalent infinitesimal), but I can't because the integral is improper of second kind: $\frac{\log(1+t)-at}{t}$ is not continuous at $t=0$.
I ran out of ideas, so any help would be appreciate. Thanks!
 A: Let$$\varphi(t)=\begin{cases}\frac{\log(1+t)-at}t&\text{ if }t\ne0\\1-a&\text{ if }t=0.\end{cases}$$Then $\varphi$ is continuous. Now, let$$F(x)=\int_0^x\varphi(t)\,\mathrm dt.$$Since $\varphi$ is continuous, $F'=\varphi$. So,$$F'(x)=\varphi(x)=1-a-\frac x2+\frac{x^2}3+\cdots$$and so, since $F(0)=0$,$$F(x)=(1-a)x-\frac{x^2}4+\frac{x^3}9+\cdots$$It follows from this that$$\int_0^{x^2}\frac{\log(1+t)-at}t\,\mathrm dt=(1-a)x^2-\frac{x^4}4+\frac{x^6}9+\cdots$$On the other hand,$$\left(1-\cos\left(\frac x2\right)\right)^2=\frac{x^4}{64}-\frac{x^6}{1536}+\cdots$$and so your limit exists if and only if $a=1$.
A: $$\lim_{x\to0}\int_0^{x^2}\frac{\log(1+t)-at}{t}dt$$.
So, this limit approaches $0$ from positive side(because $x^{2}$ keeps it so). We can find the behaviour of function as soon as it comes in the region $(0,1]$. I will do it by Taylor series expansion of $\ln(1+x)$ for $|x|<1$.
So assuming this I will directly put this Taylor series expansion inside the integral.
$$\lim_{x\to 0}\int_{0}^{x^{2}}\frac{\ln(1+t)-at}{t}dt$$
In this limit we are allowed to split the integral at the minus sign.
$$\lim_{x\to 0}\int_{0}^{x^{2}}\frac{\ln(1+t)}{t}dt -ax^{2}$$
So note that
$$\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^{k}}{k}$$
$$\lim_{x\to 0}\int_{0}^{x^{2}}\sum_{k=1}^{\infty}(-1)^{k+1}\frac{t^{k}}{tk}dt -ax^{2}$$
Bringing the summation sign and constants outside,
$$\lim_{x\to 0}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k}\int_{0}^{x^{2}}t^{k-1}dt -ax^{2}$$
$$\lim_{x\to 0}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k}\frac{x^{2k}}{k}-ax^{2}$$
Which is obviously 0, but I believe one can treat the integral like this. It eliminates usage of FTC but it works as far I am concerned.
