There is a lot to unpack here, and probably this is a better question for math.SE. TL; DR version is: in exact arithmetic, this does not converge; see Maxim Umansky's answer. In FP arithmetic, it will converge. I don't know to what, see the long version for an attempt.

Let's assume that it is possible to compute $\int_0^{\pi} f_n(x)\text{d}x$ exactly. I don't want to deal with the extra arithmetic coming from there, and it doesn't change the final answer much anyways.

After a while $\int_0^{\pi} f_n(x)\text{d}x$ will become small enough that $$fl(\sum_{n=1}^N \int_0^{\pi} f_n(x) \text{d}x+ \int_0^{\pi} f_{N+1}(x)\text{d}x)= fl(\sum_{n=1}^N \int_0^{\pi} f_n(x)\text{d}x)$$ where $fl$ is the rounding function. This is a common phenomenon in computational mathematics that you should be careful about.

(A well-known example is $\sum_{n=1}^{\infty} 1/n$ which diverges in exact arithmetic, but converges in FP arithmetic. If you reverse the ordering of the sum $\sum_{n=M}^{1} 1/n$ for $M\gg 1$, then you get the expected divergence result.)

Also, $f_n(x)$ itself can be thought as a fixed point iteration, and the integral $\int_0^{\pi} f_n(x)$ can be interpreted as testing the convergence of the fixed point iterations for all the initial guesses in the interval $[0,\pi]$ at the $n$-th iteration. The fixed point iteration $x=\sin(x)$ has the solution $x=0$ for any initial guess $x_0\in [0,\pi]$ (not too hard to prove, so I will skip). Hence, the integral $\int_0^{\pi} f_n(x)\text{d}x \to 0$ as $n\to\infty$. The upper bound for this integral is $\pi f_n(\pi/2)$.

Now, we can naively bound $\sin^n(\pi/2)$ ($\sin^n$ means $\sin$ composited with itself $n$ times) from below by $1/n$. Here is a proof by induction:

$n=1$ case: $\sin^1(\pi/2) \geq 1/1$. Hence, we are good here.

Inductive hypothesis: Assume for $n=k$, $\sin^n(\pi/2)\geq 1/n$.

$n=k+1$ case: Consider $\begin{align}\sin(\sin^k(\pi/2)) - 1/(k+1) &= \sin(\sin^k(\pi/2)) - 1/k + 1/(k^2+k) \\ &\text{by the inductive hypothesis and since sine is an increasing func. in } [0,1]\\ &\geq \sin(1/k) - 1/k + 1/(k^2+k) \\ &\text{by Taylor's remainder theorem}\\ &\geq 1/k -1/(6k^3) -1/k +1/(k^2+k)\\ &\text{for any } k\geq 1\\ &\geq 0. \end{align}$

So, the lower bound for the sum $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ is $\pi\sum_{n=1}^{\infty} 1/n$, which will definitely not increase after $n=4.5\times 10^{15}\approx 1/\varepsilon_{\text{mach}}$ since $fl(1+\varepsilon_{\text{mach}}) = fl(1)$. I would expect $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ to "converge" earlier than that (the word "converge" is in quotation marks, because it is not a real convergence).

On the other hand, we can use $1$ as an upper bound to the upper bound $f_n(\pi/2)$ (since the fixed point iterations converge to zero). Then we see that $$4.5\pi\times 10^{15}\approx\pi\sum_{n=1}^{\infty} 1 \geq \pi\sum_{n=1}^{\infty} f_n(\pi/2) \geq \pi\sum_{n=1}^{\infty} 1/n \approx \pi(\ln(4.5\times 10^{15}) + \gamma),$$ where $\gamma$ is the Euler–Mascheroni constant. (Approximation signs in the equation above mean floating point approximation, and it is non-standard notation. I am abusing the notation here)

Hence, the sum $\sum_{n=1}^{\infty} \int_0^{\pi} f_n(x)$ definitely has a finite limit when calculated on a computer using floating point arithmetic and that value is no larger than $4.5\pi\times 10^{15}$. I would expect it to be slightly larger $\pi\ln(4.5\times 10^{15})$, but I cannot prove that rigorously. That is more of an educated guess.

There is a lot to unpack here, and probably this is a better question for math.SE. TL; DR version is: in exact arithmetic, this does not converge; see Maxim Umansky's answer. In FP arithmetic, it will converge. I don't know to what, see the long version for an attempt.

Let's assume that it is possible to compute $\int_0^{\pi} f_n(x)\text{d}x$ exactly. I don't want to deal with the extra arithmetic coming from there, and it doesn't change the final answer much anyways.

After a while $\int_0^{\pi} f_n(x)\text{d}x$ will become small enough that $$fl(\sum_{n=1}^N \int_0^{\pi} f_n(x) \text{d}x+ \int_0^{\pi} f_{N+1}(x)\text{d}x)= fl(\sum_{n=1}^N \int_0^{\pi} f_n(x)\text{d}x)$$ where $fl$ is the rounding function. This is a common phenomenon in computational mathematics that you should be careful about.

(A well-known example is $\sum_{n=1}^{\infty} 1/n$ which diverges in exact arithmetic, but converges in FP arithmetic. If you reverse the ordering of the sum $\sum_{n=M}^{1} 1/n$ for $M\gg 1$, then you get the expected divergence result.)

Also, $f_n(x)$ itself can be thought as a fixed point iteration, and the integral $\int_0^{\pi} f_n(x)$ can be interpreted as testing the convergence of the fixed point iterations for all the initial guesses in the interval $[0,\pi]$ at the $n$-th iteration. The fixed point iteration $x=\sin(x)$ has the solution $x=0$ for any initial guess $x_0\in [0,\pi]$ (not too hard to prove, so I will skip). Hence, the integral $\int_0^{\pi} f_n(x)\text{d}x \to 0$ as $n\to\infty$. The upper bound for this integral is $\pi f_n(\pi/2)$.

Now, we can naively bound $\sin^n(\pi/2)$ ($\sin^n$ means $\sin$ composited with itself $n$ times) from below by $1/n$. Here is a proof by induction:

$n=1$ case: $\sin^1(\pi/2) \geq 1/1$. Hence, we are good here.

Inductive hypothesis: Assume for $n=k$, $\sin^n(\pi/2)\geq 1/n$.

$n=k+1$ case: Consider $\begin{align}\sin(\sin^k(\pi/2)) - 1/(k+1) &= \sin(\sin^k(\pi/2)) - 1/k + 1/(k^2+k) \\ &\text{by the inductive hypothesis and since sine is an increasing func. in } [0,1]\\ &\geq \sin(1/k) - 1/k + 1/(k^2+k) \\ &\text{by Taylor's remainder theorem}\\ &\geq 1/k -1/(6k^3) -1/k +1/(k^2+k)\\ &\text{for any } k\geq 1\\ &\geq 0. \end{align}$

So, the lower bound for the sum $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ is $\pi\sum_{n=1}^{\infty} 1/n$, which will definitely not increase after $n=4.5\times 10^{15}\approx 1/\varepsilon_{\text{mach}}$ since $fl(1+\varepsilon_{\text{mach}}) = fl(1)$. I would expect $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ to "converge" earlier than that (the word "converge" is in quotation marks, because it is not a real convergence).

On the other hand, we can use $1$ as an upper bound to the upper bound $f_n(\pi/2)$ (since the fixed point iterations converge to zero). Then we see that $$4.5\pi\times 10^{15}\approx\pi\sum_{n=1}^{\infty} 1 \geq \pi\sum_{n=1}^{\infty} f_n(\pi/2) \geq \pi\sum_{n=1}^{\infty} 1/n \approx \pi(\ln(4.5\times 10^{15}) + \gamma),$$ where $\gamma$ is the Euler–Mascheroni constant. (Approximation signs in the equation above mean floating point approximation, and it is non-standard notation. I am abusing the notation here)

Hence, the sum $\sum_{n=1}^{\infty} \int_0^{\pi} f_n(x)$ definitely has a finite limit when calculated on a computer using floating point arithmetic and that value is no larger than $4.5\pi\times 10^{15}$. I would expect it to be slightly larger $\pi\ln(4.5\times 10^{15})$, but I cannot prove that rigorously. That is more of an educated guess.

Edit: If Maxim Umansky's answer is not clear to you, you can look for a lower bound for the integral $\int_{0}^{\pi} f_n(x)\text{d}x$ using quadrature rules. A simple to prove lower bound comes from the observation that $f_n$ is concave (see Theorems 1 and 2) for all finite $n$ so $\int_{0}^{\pi} f_n(x)\text{d}x \geq f_n(\pi/2)\pi/2 \geq \pi/(2n)$. The first inequality basically says that the composite trapezoidal rule applied to the integral using the subintervals $[0,\pi/2]$ and $[\pi/2,\pi]$ will be a lower bound to the value of the integral. Then using the crude lower bound to $f_n(\pi/2)$, we can say that $\int_{0}^{\pi} f_n(x)\text{d}x\geq \frac{\pi}{2} \frac{1}{n}$. Hence, $\sum_{n=1}^{\infty}\int_{0}^{\pi} f_n(x)\text{d}x\geq \frac{\pi}{2} \sum_{n=1}^{\infty}\frac{1}{n}$ which diverges.

Theorem 1: If $h(x)$ is concave and non-decreasing, and $g(x)$ is concave, then $f(x) = h(g(x))$ is concave.

Theorem 2: If $h(x)$ is concave and non-increasing, and $g(x)$ is concave, then $f(x) = h(g(x))$ is concave.

Proof (example): https://math.stackexchange.com/questions/2307500/composition-of-functions-and-concavity

There is a lot to unpack here, and probably this is a better question for math.SE. TL; DR version is: in exact arithmetic, this does not converge; see Maxim Umansky's answer. In FP arithmetic, it will converge. I don't know to what, see the long version for an attempt.

Let's assume that it is possible to compute $\int_0^{\pi} f_n(x)\text{d}x$ exactly. I don't want to deal with the extra arithmetic coming from there, and it doesn't change the final answer much anyways.

After a while $\int_0^{\pi} f_n(x)\text{d}x$ will become small enough that $$fl(\sum_{n=1}^N \int_0^{\pi} f_n(x) \text{d}x+ \int_0^{\pi} f_{N+1}(x)\text{d}x)= fl(\sum_{n=1}^N \int_0^{\pi} f_n(x)\text{d}x)$$ where $fl$ is the rounding function. This is a common phenomenon in computational mathematics that you should be careful about.

(A well-known example is $\sum_{n=1}^{\infty} 1/n$ which diverges in exact arithmetic, but converges in FP arithmetic. If you reverse the ordering of the sum $\sum_{n=M}^{1} 1/n$ for $M\gg 1$, then you get the expected divergence result.)

Also, $f_n(x)$ itself can be thought as a fixed point iteration, and the integral $\int_0^{\pi} f_n(x)$ can be interpreted as testing the convergence of the fixed point iterations for all the initial guesses in the interval $[0,\pi]$ at the $n$-th iteration. The fixed point iteration $x=\sin(x)$ has the solution $x=0$ for any initial guess $x_0\in [0,\pi]$ (not too hard to prove, so I will skip). Hence, the integral $\int_0^{\pi} f_n(x)\text{d}x \to 0$ as $n\to\infty$. The upper bound for this integral is $\pi f_n(\pi/2)$.

Now, we can naively bound $\sin^n(\pi/2)$ ($\sin^n$ means $\sin$ composited with itself $n$ times) from below by $1/n$. Here is a proof by induction:

$n=1$ case: $\sin^1(\pi/2) \geq 1/1$. Hence, we are good here.

Inductive hypothesis: Assume for $n=k$, $\sin^n(\pi/2)\geq 1/n$.

$n=k+1$ case: Consider $\begin{align}\sin(\sin^k(\pi/2)) - 1/(k+1) &= \sin(\sin^k(\pi/2)) - 1/k + 1/(k^2+k) \\ &\text{by the inductive hypothesis and since sine is an increasing func. in } [0,1]\\ &\geq \sin(1/k) - 1/k + 1/(k^2+k) \\ &\text{by Taylor's remainder theorem}\\ &\geq 1/k -1/(6k^3) -1/k +1/(k^2+k)\\ &\text{for any } k\geq 1\\ &\geq 0. \end{align}$

So, the lower bound for the sum $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ is $\pi\sum_{n=1}^{\infty} 1/n$, which will definitely not increase after $n=4.5\times 10^{15}\approx \varepsilon_{\text{mach}}$ since $fl(1+\varepsilon_{\text{mach}}) = fl(1)$. I would expect $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ to "converge" earlier than that (the word "converge" is in quotation marks, because it is not a real convergence).

On the other hand, we can use $1$ as an upper bound to the upper bound $f_n(\pi/2)$ (since the fixed point iterations converge to zero). Then we see that $$4.5\pi\times 10^{15}\approx\pi\sum_{n=1}^{\infty} 1 \geq \pi\sum_{n=1}^{\infty} f_n(\pi/2) \geq \pi\sum_{n=1}^{\infty} 1/n \approx \pi(\ln(4.5\times 10^{15}) + \gamma),$$ where $\gamma$ is the Euler–Mascheroni constant.

Hence, the sum $\sum_{n=1}^{\infty} \int_0^{\pi} f_n(x)$ definitely has a finite limit when calculated on a computer using floating point arithmetic and that value is no larger than $4.5\pi\times 10^{15}$. I would expect it to be slightly larger $\pi\ln(4.5\times 10^{15})$, but I cannot prove that rigorously. That is more of an educated guess.

There is a lot to unpack here, and probably this is a better question for math.SE. TL; DR version is: in exact arithmetic, this does not converge; see Maxim Umansky's answer. In FP arithmetic, it will converge. I don't know to what, see the long version for an attempt.

Let's assume that it is possible to compute $\int_0^{\pi} f_n(x)\text{d}x$ exactly. I don't want to deal with the extra arithmetic coming from there, and it doesn't change the final answer much anyways.

After a while $\int_0^{\pi} f_n(x)\text{d}x$ will become small enough that $$fl(\sum_{n=1}^N \int_0^{\pi} f_n(x) \text{d}x+ \int_0^{\pi} f_{N+1}(x)\text{d}x)= fl(\sum_{n=1}^N \int_0^{\pi} f_n(x)\text{d}x)$$ where $fl$ is the rounding function. This is a common phenomenon in computational mathematics that you should be careful about.

(A well-known example is $\sum_{n=1}^{\infty} 1/n$ which diverges in exact arithmetic, but converges in FP arithmetic. If you reverse the ordering of the sum $\sum_{n=M}^{1} 1/n$ for $M\gg 1$, then you get the expected divergence result.)

Also, $f_n(x)$ itself can be thought as a fixed point iteration, and the integral $\int_0^{\pi} f_n(x)$ can be interpreted as testing the convergence of the fixed point iterations for all the initial guesses in the interval $[0,\pi]$ at the $n$-th iteration. The fixed point iteration $x=\sin(x)$ has the solution $x=0$ for any initial guess $x_0\in [0,\pi]$ (not too hard to prove, so I will skip). Hence, the integral $\int_0^{\pi} f_n(x)\text{d}x \to 0$ as $n\to\infty$. The upper bound for this integral is $\pi f_n(\pi/2)$.

Now, we can naively bound $\sin^n(\pi/2)$ ($\sin^n$ means $\sin$ composited with itself $n$ times) from below by $1/n$. Here is a proof by induction:

$n=1$ case: $\sin^1(\pi/2) \geq 1/1$. Hence, we are good here.

Inductive hypothesis: Assume for $n=k$, $\sin^n(\pi/2)\geq 1/n$.

$n=k+1$ case: Consider $\begin{align}\sin(\sin^k(\pi/2)) - 1/(k+1) &= \sin(\sin^k(\pi/2)) - 1/k + 1/(k^2+k) \\ &\text{by the inductive hypothesis and since sine is an increasing func. in } [0,1]\\ &\geq \sin(1/k) - 1/k + 1/(k^2+k) \\ &\text{by Taylor's remainder theorem}\\ &\geq 1/k -1/(6k^3) -1/k +1/(k^2+k)\\ &\text{for any } k\geq 1\\ &\geq 0. \end{align}$

So, the lower bound for the sum $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ is $\pi\sum_{n=1}^{\infty} 1/n$, which will definitely not increase after $n=4.5\times 10^{15}\approx 1/\varepsilon_{\text{mach}}$ since $fl(1+\varepsilon_{\text{mach}}) = fl(1)$. I would expect $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ to "converge" earlier than that (the word "converge" is in quotation marks, because it is not a real convergence).

On the other hand, we can use $1$ as an upper bound to the upper bound $f_n(\pi/2)$ (since the fixed point iterations converge to zero). Then we see that $$4.5\pi\times 10^{15}\approx\pi\sum_{n=1}^{\infty} 1 \geq \pi\sum_{n=1}^{\infty} f_n(\pi/2) \geq \pi\sum_{n=1}^{\infty} 1/n \approx \pi(\ln(4.5\times 10^{15}) + \gamma),$$ where $\gamma$ is the Euler–Mascheroni constant. (Approximation signs in the equation above mean floating point approximation, and it is non-standard notation. I am abusing the notation here)

Hence, the sum $\sum_{n=1}^{\infty} \int_0^{\pi} f_n(x)$ definitely has a finite limit when calculated on a computer using floating point arithmetic and that value is no larger than $4.5\pi\times 10^{15}$. I would expect it to be slightly larger $\pi\ln(4.5\times 10^{15})$, but I cannot prove that rigorously. That is more of an educated guess.