Alternate Way to Find Integers in Range of given Function I encountered this question recently where 
$$f(x)=\int_0^{x} {e^{t}\sin(x-t)}$$
and 
$$g(x)=\dfrac{d^{2}f(x)}{dx^{2}} - f(x)$$
I had to find the number of integers in the range of $~g(x)~$
I solved it analytically by computing it's second derivative after substituting $$x=x-t$$ and computing the function $~g(x)~$ which came out to be $\sin x+\cos x$.
However I was wondering is there a way to do this without actually computing the derivative and using some sort of numerical analysis or such technique to check for Integers in the range without using any significant computation power.
 A: Actually, that is possible. And I also found that your answer has a small problem.
$$f(x)=\int_0^{x} {e^{t}\sin(x-t)}=-e^{t}\cos(x-t)+e^t\cos(-t)$$
When we take the second derivative of $f(x)$, $e^t\cos(-t)$ becomes $0$. So we don't consider this constant until the end, because any vertical translation applied on the function will only translate its range.So, let $$y(x)=-e^{t}\cos(x-t)$$
Instead of calculating the $2^{nd}$ derivative, we use a general pattern for this kind of trigonometric functions: Its $2^{nd}$ derivative is equal to $-f(x)$ (Note that I didn't calculate it)
Hence, $g(x)=-2y(x)-C$ where $C=e^t\cos(-t)$, but please ignore $C$.
Now, found that any stationary point of $y(x)$ appear when $\cos(x-t)$ reaches its minimum or maximum. Therefore, we require:
$$x-t=k\pi, k\in\mathbb{N}$$
Since $y(x)$ is periodic, we only need to investigate one pair of points, then $x=k\pi+t$, plug this result in $y(x)$ you'll get:
$$y(k\pi+t)=-e^{t}\cos(k\pi)$$
Apply $-2y(x)$, The number of integers:$$N=\lfloor|2e^{t}\cos(k\pi)|-C\rfloor+|\lceil-|2e^{t}\cos(k\pi)|-C\rceil |+1$$
The reason for $+1$ is that we need to count $0$.
