# Closed form of $\int_0^\pi \frac{\sin(x)}{\sqrt{x^3+x+1}} dx$

I'm looking for a closed-form expression for the value of this integral:

$$I=\int_0^\pi \frac{\sin(x)}{\sqrt{x^3+x+1}} dx$$

The graph of the integrand looks like this:

$\hskip 2.4 in$

Numerically, the area is $0.875044...$ for which the Inverse Symbolic Calculator doesn't produce anything promising. My CAS finds neither an antiderivative nor a closed form for the definite integral, and my own manipulations haven't really got me anywhere either.

• i also think there is no closed form for this integral, use a numerical method – Dr. Sonnhard Graubner Oct 21 '14 at 6:27
• This is the mixed algebraic-transcendental case for the Risch algorithm, which I don't think any of the usual CAS's have implemented fully. However, it's probably true that there is no closed-form antiderivative. Yes, the fact that it's a definite integral does give some hope, but I don't immediately see any way to do it using residues etc. – Robert Israel Oct 21 '14 at 6:27
• The real challenge lies within the integral $\int \sqrt {x^3 + x + 1}$ , if we can find even an approximate closed form of this integral our problem is solved. – Awe Kumar Jha Oct 28 '18 at 5:04

# A Neater Expression

$$I=\int^π_0 \frac {\sin x}{\sqrt {x^3+x+1}} dx = \sum_0^∞ A_k \sum_0^k (-1)^r {}^kP_{2r} π^{k-2r}$$ Where, $$(3+2k)A_k + (5+2k)A_{2+k} + 2(3+k)A_{3+k} = 0,$$ $$A_0=1, A_1=\frac {-1}{2}, A_2=\frac {3}{8}$$

# A greedy approach

$$I=\int^π_0 \frac {\sin x}{\sqrt {x^3+x+1}} dx = 0.8750439062939084$$ Using the greedy Egyptian fraction algorithm, $$x_{k+1} = x_k - \frac {1}{\lceil \frac {1}{x_k} \rceil}$$ where, $$x_0 = I$$, I got an expansion, $$I = \frac {1}{2} + \frac {1}{3} + \frac {1}{2^3.3} +\frac {1}{2^3.3.13.73}+\frac {1}{2^2.13.113.397547} +……$$ I couldn't go farther , for my limited computational capacity (which is my laptop), however I indeed see one pattern : the prime factors in the denominators $$(2,3,13,73,113,…)$$ belong to the set of primes given by, $$a(n)= \text {Min} \left(x; π[x]-π\left[\frac {x}{2}\right]=n\right)$$ I got it on OEIS(https://oeis.org/A080359). Yet it needs much more insight.

A closed form would be extremely difficult to get. This appears to be a new function. Substituting $$t$$ for the denominator, we get a beautiful form of the integral, however potentially latent in the present context. $$I = 2\int \frac {\cosh J(t)}{\cosh 3J(t)} \sin \left(-2 \sqrt {\frac {1}{3}} \sinh J(t)\right) dt$$ where, $$J(t):=\frac {1}{3} \sinh^{-1} \left[\frac {3\sqrt 3}{2} (1-t^2)\right]$$ So , I am giving a series form solution. Consider, $$F(x):=\int \frac {\sin x}{\sqrt {1+x(1+x^2)}} dx\tag{1}$$ Now, for $$x<1$$, $$[1+x(1+x^2)]^{-\frac {1}{2}} = -\sum_{k=0}^∞ C^{k-\frac {1}{2}}_{-\frac {1}{2}}x^k(1+x^2)^k \tag{2}$$ Plugging $$(2)$$ into $$(1)$$ we get, $$F(x) = -\sum_{k=0}^∞ C^{k-\frac {1}{2}}_{-\frac {1}{2}} G(k,x) \tag{3},$$ where , $$G(k,x):= \int x^k \sin x (1+x^2)^k dx \tag{4}$$ But, $$(1+x^2)^k= \sum_{r=0}^k C^k_r x^{2r} \tag{5}$$ Plugging $$(5)$$ into $$(4)$$ we get, $$G(k,x) = \sum_{r=0}^k C^k_r H(r,k,x) \tag{6},$$ where, $$H(r,k,x) := \int x^{k+2r} \sin x dx$$ $$= - \frac {\Gamma (k+2r+1, ix) + (-1)^{k+2r}\Gamma (k+2r+1, -ix)}{2(-1)^{\frac {5}{2} (k+2r)}} \tag{7}$$ Hence, $$F(x)=\sum_{k=0}^∞ C^{k-\frac {1}{2}}_{-\frac {1}{2}} \sum_{r=0}^k C^k_r \frac {\Gamma (k+2r+1, ix) + (-1)^{k+2r}\Gamma (k+2r+1, -ix)}{2(-1)^{\frac {5}{2} (k+2r)}}\tag{8}$$ On the same lines an integral exists for the case $$x>1$$, the only difference being in the binomial expansion for the denominator of the original integral . Call it $$F'(x)$$. Then, $$\int_0^π \frac {\sin x}{\sqrt {1+x(1+x^2)}} dx = [F(1) -F(0)] +[F'(π) - F'(1)]$$ This solution is in terms of upper incomplete gamma functions with complex arguments. The notation $$C^n_r$$ stands for combinatorial coefficients.
Alternatively, one could use hypergeometric functions to express the final result, $$F(x)=-\sum_{k=0}^∞ C^{k-\frac {1}{2}}_{-\frac {1}{2}} \sum_{r=0}^k C^k_r \frac {x^{k+2r+2}}{k+2r+2} {}_1\text {F}_2 \left(\frac {k+2r+4}{2},\frac {3}{2} ; \frac {k+2r+2}{2} ; -\frac {x^2}{4}\right)$$
There is still hope for a closed form of the indefinite integral in terms of the Fresnel integral $$C(z)$$, consider a related integral: $$\int \frac {\sin x}{\sqrt {x^3}} dx = \sqrt {8π} \text {C} \left(\sqrt {\frac {2x}{π}}\right) - \frac {2\sin x}{\sqrt x} + c$$ The integrand is almost the same, except for the extra linear term $$x+1$$ under square root.