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

A somewhat common trick when evaluating integrals is to add a second (or third) integral to help integrate the first. As an example if one where to compute $$I = \int \sin \log x \, \mathrm{d}x\,.$$ A clever step would be to introduce the integral $$J = \int \cos \log x \, \mathrm{d}x\,,$$ and to turn the problem into a system of equations. Since now it is easy to show by parts that \begin{align*} I + J & = \left[ x \sin \log x - J \right] + J \\ J - I & = \left[ x \cos \log x + I \right] - I \end{align*} One can now solve this system of equation with ease. As one can see to integrate $I$ or $J$ by itself is somewhat cumbersome (One can use the substitution $t = \log x$, and then rewrite the remaining expression into a complex one, or use more integration by parts, but both methods are somewhat tedious), but by looking at their products or difference the computation can be simplified.

Another example are the integrals $$I = \int \frac{\cos x}{\cos x + \sin x}\,\mathrm{d}x \quad \text{and} \quad J = \int \frac{\sin x}{\cos x + \sin x}\,\mathrm{d}x$$ where the same technique can be applied.

My question is: Are there any more examples of integrals that can be solved by transforming them into a system of equations?

(please do not say $e^x \cos x$, since this is the same as the first example) Any suggestions,examples, reference to books, papers, is much appriciated.

## 4 Answers

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

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

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

This is an example such that adding the two makes it easier to solve though solving each integral seems difficult.

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.

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 $$\frac{1}{2}x\sin\log x - \frac{1}{2}x\cos\log x.$$

Here's an article about applications of linear algebra to calculus. Essentially what you're doing is using those functions as the basis to a vector space.

http://www.jstor.org/discover/10.2307/2974819?uid=3739920&uid=2134&uid=2473376467&uid=2&uid=70&uid=3&uid=2473376457&uid=3739256&uid=60&purchase-type=article&accessType=none&sid=21102532976847&showMyJstorPss=false&seq=4&showAccess=false

So to approach the problem again in these new terms, $e^xsinx$ and $e^xcosx$ are your basis vectors which have a derivative matrix which maps back onto itself. The inverse of your derivative matrix is the integral matrix, so you can just do a simple vector times your matrix to find the answer of any linear combination of your two basis vectors.

The main thing is that they're connected one to the other by differentiation or integration. Your method involves using integration by parts, but this method simply involves the derivative.

For any power or fourier series you could always try inverting an infinite matrix with a basis of 1, x, x^2, etc... to integrate anything after you've expressed it as such.