Series Representation of the Glasser function: $\text G(x)\mathop=\limits^\text{def} \int_0^x \sin(t\sin(t))dt\sim2\sqrt{\frac x\pi}$ Here is an uncommon special function called the Glasser function as referenced by Wolfram Mathworld.
which is defined as:
$$\text G(x)\mathop=^\text{def} \int_0^x \sin(t\sin(t))dt\sim2\sqrt{\frac x\pi}$$
Here is an interactive graph of the function:

One idea for the evaluation for the function is:
$$\int_0^x \sin(t\sin(t))dt=\int_0^x \sum_{n=0}^\infty \frac{(-1)^n t^{2n+1}\sin^{2n+1}(t)}{n!}dx$$
which could almsot be expressed in terms of Generalized Log-Sine Integral function, but this does not seem to be able to be put in closed form.
Let’s see what happens when we do a substitution of $$t=\sin^{-1}(y),y=\sin(t)\implies dt=\frac{dy}{\sqrt{1-y^2}},y_1=\sin(0)=0,y_2=\sin(x)$$
$$\int_0^x\sin(t\sin(t))dt=\int_0^{\sin(x)}\frac{\sin\left(y\sin^{-1}(y)\right)}{\sqrt{1-y^2}}dy$$
Which reminds us about the Central ChebyshevU function:
$$\text U_t(t)=\frac{\sin\left((t+1)\cos^{-1}(t)\right)}{\sqrt{1-t^2}}$$
which just may put the integrand into a simpler form.
The function comes from just one source which I cannot find online:

Glasser, M. L. and Cosgrove, C. "Problem $785$." Nieuw Arch. Wisk. $8, 455, 1990$.

One possible solution is via Incomplete Bessel function:
$$\int_0^x \sin(t\sin(t))dt=\text{Im}\int_0^x e^{it\sin(t)}dt$$
What is a sum representation of the integral? The integral should be a single sum with a closed form summand, so no nth derivatives please. A closed form is also appreciated if possible, but is optional. Please correct me and give me feedback!
 A: The series representation for
$$I(x):=\int_0^x\sin(t\ \sin{t})dt $$
is
$$ I(x)=4\sum_{j=0}^\infty \frac{(x/2)^{2j+3}}{2j+3} \sum_{k=0}^j (-1)^k \sum_{m=0}^k
\frac{( 2(2(j-k)+1))^{2k+1-2m}}{m!(m+2(j-k)+1)!(2k+1-2m)!} $$
A proof-sketch is as follows.  From Gradshteyn and Ryzhik 8.511.3 one derives a Bessel function representation
$$\sin(t \, \sin{t})=2\sum_{k=0}^\infty J_{2k+1}(t) \sin\big((2k+1)t \big).$$
By a Cauchy product, and with some algebra one derives
$$ J_{2k+1}(t) \sin\big((2k+1)t \big) = (t/2)^{2k+1} \sum_{n=0}^\infty (-1)^n(t/2)^{2n+1} \sum_{m=0}^n \frac{ (2(2k+1))^{2n+1-2m}}{m!(m+2k+1)!(2n+1-2m)!} $$
The integration over t is now easy, and we get the double sum
$$I(x)=4(x/2)^3 \sum_{n,k=0}^\infty \frac{(x/2)^{2(n+k)}}{2(n+k)+3}(-1)^n 
\sum_{m=0}^n \frac{ (2(2k+1))^{2n+1-2m}}{m!(m+2k+1)!(2n+1-2m)!} $$
Use a series rearrangement with $n+k=j$
$$I(x)=4(x/2)^3 \sum_{j=0}^\infty \frac{(x/2)^{2j}}{2j+3}\sum_{k=0}^{j}(-1)^{j-k} 
\sum_{m=0}^{j-k} \frac{ (2(2k+1))^{2(j-k)+1-2m}}{m!(m+2k+1)!(2(j-k)+1-2m)!} $$
Summing over the innermost sum in $k$ but in the reverse direction gives the form at the top of the answer.  I have not proved the radius of convergence for the series, so it should be thought of as asymptotic for $x \to 0$ until then.  I've checked a few values up to $x=3.1,$ and get decent numerical agreement.  For $x=3.1,$ I truncated the outer sum at j=200, the absolute value of summand is about $3 \cdot 10^{-10}$ and the numerical integration vs the truncated sum agree to about 6 digits precision.
A: Here is a recursive algorithm to generate the coefficients of the Maclaurin series of $G(x)$. Set
$$\tag{1}
\sin (x\sin x) = \sum\limits_{k = 1}^\infty  {a_k x^{2k} } 
$$
and
$$\tag{2}
\cos (x\sin x) = \sum\limits_{k = 0}^\infty  {b_k x^{2k} } .
$$
Differentiating each side of $(1)$ gives
$$
\cos (x\sin x)(\sin x + x\cos x) = \sum\limits_{k = 0}^\infty  {(2k + 2)a_{k + 1} x^{2k + 1} } 
$$
or
$$
\left( {\sum\limits_{k = 0}^\infty  {b_k x^{2k} } } \right)\left( {\sum\limits_{k = 0}^\infty  {( - 1)^k \frac{{2k + 2}}{{(2k + 1)!}}x^{2k + 1} } } \right) = \sum\limits_{k = 0}^\infty  {(2k + 2)a_{k + 1} x^{2k + 1} } .
$$
Dividing through by $x$ and performing the Cauchy product on the left-hand side yields the recurrence
$$\tag{3}
a_{n + 1}  = \frac{1}{{2n + 2}}\sum\limits_{k = 0}^n {( - 1)^k \frac{{2k + 2}}{{(2k + 1)!}}b_{n - k} }
$$
for $n\geq 0$ with $b_0=1$. Similarly, differentiating both sides of $(2)$ leads to the recurrence
$$\tag{4}
b_{n + 1}  =  - \frac{1}{{2n + 2}}\sum\limits_{k = 0}^{n - 1} {( - 1)^k \frac{{2k + 2}}{{(2k + 1)!}}a_{n - k} } 
$$
for $n\geq 1$ with $a_1=1$. The recurrences $(3)$ and $(4)$ together with the initial values $a_1=1$ and $b_0=1$ allow us to generate all the coefficients $a_n$. Finally, integrating $(1)$,
$$
G(x) = \int_0^x {\sin (t\sin t)dt}  = \sum\limits_{n = 1}^\infty  {\frac{{a_n }}{{2n + 1}}x^{2n + 1} } 
$$
for all $x \in \mathbb{R}$.
A: I was experimenting and found these interesting integral representations with a power tower like integrand:
$$\text G(x)=\text{Im}\int_0^xe^{it\sin(t)}dt\mathop=^{t=\tan^{-1}(T)}\text{Im}\int_0^{\tan(x)}\frac{e^{i\tan^{-1}(T)\sin(\tan^{-1}(T))}}{T^2+1}dT= \text{Im}\int_0^{\tan(x)}\frac{\left(\frac{1+ iT}{\sqrt{T^2+1}}\right)^\frac T{\sqrt{T^2+1}}}{T^2+1}dT\mathop=^{T^2+1=y} \frac 12\text{Im}\int_1^{\sec^2(x)}\frac{\left(\frac 1{\sqrt y}+i\sqrt{1-\frac1y}\right)^\sqrt{1-\frac1y}}{\sqrt{y(y-1)}}\mathop=^{1-\frac 1y=v} \frac12\text{Im}\int_0^{\sin^2(x)}\frac{\left(i\left(\sqrt{v-1}+\sqrt{v}\right)\right)^\sqrt v}{|v-1|\sqrt v}dv$$
and:
$$\text G(x)=\text{Im}\int_0^xe^{it\sin(t)}dt\mathop=^{t=\ln(T)}\text{Im}\int_1^{e^x} e^{i\ln(T)\sin(\ln(T))}T^{-1}dT=\text{Im}\int T^{\frac{T^i-T^{-i}}{2}-1}dT\mathop=^{T=y^i}i\text{Im}\int_1^{e^{-i x}}y^{i\frac{\frac 1y-y}2+i-2}dy$$
So we have this closed for with a function from Wolfram Mathworld, but these forms are hard to use. Please correct me and give me feedback!
