Determine $a, b, c, d$, where $\int_0^x{\frac{t}{\arctan(t)}}dt=a+bx+cx^2+dx^3+o(x^3)$

Question

Let $$f(x)=\int_0^x{\frac{t}{\arctan(t)}}dt=a+bx+cx^2+dx^3+o(x^3)$$ be the function defined by the above integration where $x \not= 0$.

The exercise is to determine the constants $a, b, c, d$, but I was unable to develop the solution. The $o(x^3)$ is the o notation for the error in Taylor polynomials.

Answer 1

Hint: Note that $a_n=\dfrac{f^{(n)}(0)}{n!}$ $$a=f(0)$$ $$b=f'(0)=\lim_{x\to0}f'(x)$$ $$c=\frac12f''(0)=\frac12\lim_{x\to0}f''(x)$$ and so on

Answer 2

$$\frac t{\arctan t} = 1 + t + \frac 13t^2 + 0t^3 - \frac 4{45}t^4 + 0t^5 + O(t^6)$$ $$$$ $$\int_0^x {(1 + t + \frac 13t^2 + 0t^3 - \frac 4{45}t^4 + 0t^5)} dt$$ $$ = (t + \frac 12t^2 + \frac 1{3*3}t^3 + 0t^4 - \frac 4{45*5}t^5 + 0t^6)\bigg|_0^x$$ $$= 0 + x + 0x^2 + \frac19x^3 + 0x^4 -\frac 4{225}x^5$$

Answer 3

Use the standard series $$\tan ^{-1}(t)=t-\frac{t^3}{3}+\frac{t^5}{5}+O\left(t^{7}\right)$$ $$\frac{t}{\tan ^{-1}(t)}=\frac {t} {t-\frac{t^3}{3}+\frac{t^5}{5}+O\left(t^{7}\right) }=\frac {1} {1-\frac{t^2}{3}+\frac{t^4}{5}+O\left(t^{6}\right) }$$ Use long division $$\frac{t}{\tan ^{-1}(t)}=1+\frac{t^2}{3}-\frac{4 t^4}{45}+O\left(t^6\right)=1+\frac{t^2}{3}-\frac{4 t^4}{45}+O\left(t^6\right)$$ $$\int \frac{t}{\tan ^{-1}(t)}\,dt=t+\frac{t^3}{9}-\frac{4 t^5}{225}+O\left(t^{7}\right)$$ Remember that $\frac{t}{\tan ^{-1}(t)}$ is even and, so, the integral is odd; so the coefficients for even power of $t$ are zero.

Answer 4

Letting $x\to0$ gives $a=0$ since $\lim_{t\to0}\frac{t}{\arctan t}=1$. Taking derivative for both sides gives $$ x=\bigg[b+2cx+3dx^2+O(x^3)\bigg]\arctan x. $$ Using $$ \arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}+O\left(t^{7}\right) $$ one has $$ x=\bigg[b+2cx+3dx^2+O(x^3)\bigg]\bigg[x-\frac{x^3}{3}+\frac{x^5}{5}+O\left(t^{7}\right)\bigg]. $$ Comparing the coefficients of $x$, $x^2$ and $x^3$, respectively, one obtains $$ b=1, c=0, -\frac b3+3d=0 $$ and hence $$ a=0,b=1,c=0,d=\frac19. $$