Prove that $\frac{x^{i+1}}{-2+2i}-\frac{x^{-i+1}}{2+2i}=\frac{x\sin(\ln(x))-x\cos(\ln(x))}{2}$

Question

While trying to integrate $\sin(\ln(x))$, I used both Euler's formula for complex sine as well as a more traditional integration by parts method. The first method gave me the solution

$\int\frac{e^{i\ln(x)}-e^{-i\ln(x)}}{2i}\text dx=\frac{1}{2i}(\int x^i\text dx-\int x^{-i}\text dx)=\frac{1}{2i}(\frac{x^{i+1}}{i+1}-\frac{x^{-i+1}}{-i+1})=\frac{x^{i+1}}{-2+2i}-\frac{x^{-i+1}}{2+2i},$

while the second method, using the substitution $u=\ln(x)$, gave

$\int\sin(\ln(x))\text dx=\int e^u\sin(x)\text dx=e^u\sin(u)-e^u\cos(u)-\int e^u\sin(u)\text du\\ 2\int e^u\sin(u)\text du=e^u\sin(u)-e^u\cos(u)\\ \int\sin(\ln(x))\text dx=\frac{x\sin(\ln(x))-x\cos(\ln(x))}{2}$

I checked both solutions in WolframAlpha and they both appear to be correct and equivalent to each other. However, I'm not sure how to go about proving that they are equivalent. Is there any way to verify that both solutions are the same? I've tried rearranging the first solution to see if I can put it in terms of sine and cosine, but can't figure out any way to do so.

Answer 1

Note that The two fractions on the left are conjugates. $$\frac{x^{i+1}}{-2+2i}-\frac{x^{-i+1}}{2+2i}= \frac{x\sin(\ln(x))- x\cos(\ln(x))}{2}$$

Since $$Z+\overline Z =2Re(Z) $$ All we have to do is to find $$ 2Re(\frac{x^{i+1}}{-2+2i})$$

We have $$x^i = e^{i ln(x)}=cos(ln(x))+isin(ln(x))$$ and $$\frac {1}{-1+i}=\frac {-1-i}{2}$$

Upon multiplication and separation of the real part from the imaginary part we get the desired result of $$ 2Re(\frac{x^{i+1}}{-2+2i}) =\frac{x\sin(\ln(x))-x\cos(\ln(x))}{2}$$

Answer 2

On the left side multiply numerator and denominator by $(-2-2i)$ in the first term and by $(2-2i)$ in the second. On the right side use the identities $\sin x=\frac {e^{ix}-e^{-ix}} {2i}$ $\cos x=\frac {e^{ix}+e^{-ix}} {2}$. After a little simplification you will see that the two answers are the same.

Answer 3

you already have

$\sin\ln x = \frac 1{2i}(x^i - x^{-i})$

the rest of it:

$x\sin\ln x = \frac 1{2i}(x^{i+1} - x^{-i+1})\\ x\cos\ln x = \frac 1{2}(x^{i+1} + x^{-i+1})\\ \frac {x\sin\ln x - x\cos \ln x}{2} = \frac {1}{4}[-(i+1)x^{1+i}-(-i+1) x^{-i+1}]$

And $\frac {x^{1+i}}{-2+2i} - \frac {x^{1-i}}{2+2i}$ should be simplified, multiplying top and bottom of each fration by the complex conjugate of the denominator.