Example of integral pairs or triples ($I$, $J$, $K$...)

Solution 1:

I'm not quite sure if this is what you want.

$$I=\int_{\frac{\pi}{3}}^{\frac{2}{3}\pi}\frac{\cos(x)\sin(3x)}{\cos\left(x+\frac{2}{3}\pi\right)}dx,\qquad J=\int_{\frac{\pi}{3}}^{\frac{2}{3}\pi}\frac{\cos(x)\sin(3x)}{\cos\left(x-\frac{2}{3}\pi\right)}dx.$$

Letting $t=\pi-x$, we can get $I=J.$ Hence, since we know $I=\frac{I+J}{2}$, we get, setting $u=\cos x$, $$I=\frac{I+J}{2}=\cdots =\int_{0}^{\frac 12}\frac{4u^2(4u^2-1)}{3-4u^2}du=\cdots=-\frac 76+\frac{\sqrt3}{2}\ln(2+\sqrt 3).$$

This is an example such that adding the two should make it easier to find though finding each integral seems difficult.

Solution 2:

Consider

$$I=\int_{0}^{\infty}e^{-z^2}\cos\left(\frac{a^2}{z^2}\right)dz$$ and

$$J=\int_{0}^{\infty}e^{-z^2}\sin\left(\frac{a^2}{z^2}\right)dz$$ The calculation of this couple comes down to solving the following system of differential equations:

$$\frac{d^2I}{da^2}=4J$$

$$\frac{d^2J}{da^2}=-4I$$ which is easy.

Solution 3:

This isn't an answer to your question… but for a simple direct method (simpler, I would argue, than the system-of-equations approach), note that$$ \sin\log x =\frac{1}{2i}\left(e^{i\log x} - e^{-i\log x}\right)=\frac{1}{2i}\left(x^i-x^{-i}\right), $$ the integral of which is clearly $$ \frac{1}{2i}\left(\frac{x^{1+i}}{1+i}-\frac{x^{1-i}}{1-i}\right)=\frac{x}{2i}\left(\frac{x^i(1-i)}{2}-\frac{x^{-i}(1+i)}{2}\right)=\frac{x}{4i}\left(x^i-x^{-i}\right)-\frac{x}{4}\left(x^i+x^{-i}\right), $$ or $$ \frac{1}{2}x\sin\log x - \frac{1}{2}x\cos\log x. $$