Integrate sin(a * ln(x)) / x^b with respect to x, where a and b are constants.

Step-by-step solution

1. 1

   Substitute $u=\ln x$ to remove the logarithm from inside the sine. Then $x=e^{u}$ and $dx=e^{u}du$, so

   $\int\frac{\sin(a\ln x)}{x^{b}}dx=\int\frac{\sin(au)}{e^{bu}}e^{u}\,du=\int e^{(1-b)u}\sin(au)\,du$

   What was an awkward mix of a logarithm and a power has become the classic exponential-times-sine integral.

2. 2

   Recall the standard result. For constants $k$ and $a$ not both zero,

   $\int e^{ku}\sin(au)\,du=\frac{e^{ku}\left(k\sin(au)-a\cos(au)\right)}{k^{2}+a^{2}}+C$

   It is proved by integrating by parts twice and solving the resulting equation for the integral — the integral reappears on the right, which is why no third integration is needed.

3. 3

   Apply it with $k=1-b$.

   $\int e^{(1-b)u}\sin(au)\,du=\frac{e^{(1-b)u}\left[(1-b)\sin(au)-a\cos(au)\right]}{(1-b)^{2}+a^{2}}+C$

4. 4

   Substitute back, using $e^{(1-b)u}=x^{1-b}$ and $u=\ln x$.

   $\int\frac{\sin(a\ln x)}{x^{b}}dx=\frac{x^{1-b}\left[(1-b)\sin\left(a\ln x\right)-a\cos\left(a\ln x\right)\right]}{(1-b)^{2}+a^{2}}+C$

   The formula requires $(1-b)^2+a^2\ne 0$, that is, not both $a=0$ and $b=1$.

5. 5

   Check the special case and verify numerically. When $b=1$ the integral reduces to $\int\tfrac{\sin(a\ln x)}{x}dx$, and the formula gives $\tfrac{-a\cos(a\ln x)}{a^2}=-\tfrac{\cos(a\ln x)}{a}$, exactly what $u=\ln x$ gives directly $\checkmark$. Numerically, differentiating the general antiderivative at $a=1.7$, $b=0.4$ and $x=0.7,\,1.3,\,2.5,\,4.0$ reproduces the integrand to eight decimal places $\checkmark$.