Integration arrives to the original integral, why?

Question

I have this integral:

$$ \int{e^{x} \sin{\left (a - x \right )} d x}\tag{1} $$

After two consecutive integration by parts I arrived to the original integral:

$$ \int{e^{x} \sin{\left (a - x \right )} d x} = e^{x} \sin{\left (a - x \right )} + e^{x} \cos{\left (a - x \right )} - \int{e^{x} \sin{\left (a - x \right )} d x}\tag{2} $$

I know that when you arrive to the original integral you can solve it in the following way.

In (2) add the original integral (1) in both sides. You get to:

$$ 2\int{e^{x} \sin{\left (a - x \right )} d x} = e^{x} \sin{\left (a - x \right )} + e^{x} \cos{\left (a - x \right )}\tag{3} $$

Divide by two in both sides, and add the constant $C$ on right-hand side:

$$ \int{e^{x} \sin{\left (a - x \right )} d x} = \frac{e^{x} \sin{\left (a - x \right )} + e^{x} \cos{\left (a - x \right )}}{2}+C\tag{4} $$

My questions are: Why does this happen? That you arrive to the same integral that you were integrating. In the case of exponents you get something bigger ($\int{u^ndu} = \dfrac{u^{\bf{n+1}}}{n+1}+C$), but in this case I arrived to the very same function. Is it related to sine being a periodic function?

And, I solved it using integration by parts. Why is it valid to solve for the original integral?

Answer 1

One thing that is very different about the derivatives of $\sin u$ and $u^n$ is that when we repeatedly differentiate $u^n$ we get $\frac{d}{du} u^n = u^{n-1}$, $\frac{d^2}{du^2} u^n = u^{n-2}$, $\frac{d^3}{du^3} u^n = u^{n-3}$, etc., and we never "get back" to $u^n$, but the second derivative of $\sin u$ with respect to $u$ is $$ \frac{d^2}{du^2} \sin u = -\sin u. $$ That is, by differentiating twice, you get back the original function but with a sign change.

It's not merely that the sine function is periodic, but that it is the same shape as its own derivative, just a little out of phase: $$ \frac{d}{du} \sin u = \cos u = \sin\left(u +\frac\pi4\right). $$

As for why your method of solving the integral is valid: if an equation is true then it's true, and so is anything else that you can derive by valid reasoning from the equation. Notice that "getting back the same function" doesn't always get you the kind of answer you want; for example, if you apply integration by parts twice more to your integral you will get \begin{align} \int e^x \sin(a - x)\, dx &= e^x \sin(a - x) + e^x \cos(a - x) \\ & \qquad - e^x \sin(a - x) + e^x \cos(a - x) + \int e^x \sin(a - x)\, dx. \end{align} This also is a true equation, but when you try to "collect" the integrals on the left-hand side this equation tells you nothing more than the obvious fact that $$ e^x \sin(a - x) + e^x \cos(a - x) - e^x \sin(a - x) + e^x \cos(a - x) = 0. $$ In fact, the second integration by parts actually doesn't give you back exactly the integral you started with: it gives you minus that integral, which is what makes it possible to solve the integral the way you did.

Answer 2

Hint: use integration by parts twice. For some intuition, note that $\frac{d^2}{dx^2}\sin(x-a)=-\sin(x-a)$, so if you view $f(x)=\sin(x-a)$ andd $g^{\prime}(x)=e^{x}$, then $fg-\int f^{\prime}g $ [note that $g^{\prime}=g$!] should give the first step of the solution.

Answer 3

This is due to the fact that the derivatives of $e^x$ and $\sin (a + x)$ form "cycles". In the case of $y = e^x$, its derivative is itself. In the case of $y = \sin(x + a)$, we have: \begin{equation*} \begin{split} & y = \sin(x + a) \\ & y' = \cos(x + a) \\ & y'' = -\sin(x + a) \\ \end{split} \end{equation*} and we can already stop here, as $y'' = -y$. Therefore, when you take the antiderivatives of these functions, the same patterns will appear, only in the reverse order. While it may have to do with the sine function being periodic, the underlying reason your integral appears again has to do with the successive derivatives of the exponential and sine/cosine functions.

Solving for the integral in your second equation is just algebra; the value of the integral you are solving for as an algebraic expression, so you may treat it as such.

Indeed, the same pattern does not apply with polynomials: taking the derivative lowers the degree of the polynomial by $1$ each time, so there is no chance for a finite "cycle" of derivatives.