# Choosing a function so that the integral takes a particularly simple form

I am trying to model the behavior of a certain phenomena and in the course of it I have got a complicated integral which involves among other things an unknown function $f(x)$ (which I abbreviate to $f$). I have the option of choosing this function $f$ so that the integral can be evaluated easily.

Specifically, let $g(x)=(2^{36/5}(-5x-1)^{2/5}(x+1)^2)/x$ and consider the indefinite integral $\int \big (\frac{2f(1+x)}{1+5x}-\frac{1+x}{x(1+5x)}\big )g(x)dx$. I want to choose the function $f$ so that $f(0)=0$, $f$ is increasing and that the integral takes a particularly simple form in terms of elementary functions. How do I do that? I have been randomly selecting functions $f$ and trying them out but to no avail. Is there a standard way to solve such a problem?

Edit: If I let the integrand be denoted by $F$, then $f$ evaluates to $\frac{Fx(1+5x)^{3/5}}{2^{41/5}(1+x)^3}+\frac{1}{2x}$. But now there is no way I can have f(0)=0. How do I get over this problem?

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There's probably no standard way to solve such a problem, since integration is art; this is just how I would approach this particular problem.

Simplifying and dropping constant factors leads to

$$\int\frac{2xf(x+1)-(x+1)}{x^2(-5x-1)^{3/5}}(x+1)^2\mathrm dx\;.$$

The problem is that we have two different factors in the denominator. If we can choose $f$ such that it gets rid of one of them while leaving a polynomial in the numerator, we can make a linear transformation that turns the other one into a power through which we can divide the numerator. Thus we'd like to have

$$2xf(x+1)-(x+1)=x^2p(x)$$

with $p(x)$ a polynomial. Substituting $t=x+1$ and solving for $f$ yields

$$f(t)=\frac12\left((t-1)p(t-1)+\frac{t}{t-1}\right)\;.$$

This has a pole at $t=1$. You've specified a value for $t=0$ and the integrand has a singularity at $t=4/5$ and is undefined for $t\gt4/5$ since the radicand becomes negative, so I assume that you wouldn't mind a pole at $t=1$. If so, we just have to choose $p$ to fulfill the conditions on $f$. The condition $f(0)=0$ implies that $p$ vanishes at $t=0$, and thus $p(t-1)=tq(t-1)$. Other than that, we're free to choose $p$ such that $f$ is increasing. The second term in $f$ is decreasing, so the first term has to compensate for it. The first term is proportional to $t(t-1)q(t-1)$, and we must have $q(-1)\le-1$ for $f$ to be increasing at $t=0$. Since $-t(t-1)$ is decreasing for $t\gt1/2$, $-q$ needs to increase sufficiently in that range to compensate the decrease. We can't have further factors of $t$, but we can add any number of factors of $t+1$. With $q(t-1)=-2(t+1)^n$, we have

$$f(t)=-t(t-1)(t+1)^n+\frac12\frac t{t-1}\;.$$

Trial and error shows that $n=9$ is enough to make $f$ increasing up to $t=4/5$. (Here's a plot.) This corresponds to

$$p(x)=-2(x+1)(x+2)^9\;,$$

so the integral becomes (up to constant factors)

$$\int \frac{(x+1)^3(x+2)^9}{(-5x-1)^{3/5}}\mathrm dx\;.$$

Now the substitution $u=-5x-1$ leads to a trivial integral.

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Perhaps you could try $f(x)=x^c$, for some constant positive c.

I think using a computer algebra software would help.

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Working backward should work. Rather than selecting $f$ and hoping that the anti-derivative is simple, select the anti-derivative and solve for $f$.

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See my comment above. –  Shahab Mar 31 '12 at 11:09
Your expression doesn't look right. (among other issues, it looks like you calculated $f(x+1)$) But assuming it were right, that just means that near $0$, $F(x)$ has to look like $-2^{41/5} / (2x^2)$, so as to cancel out the pole that $1/(2x)$ has at $0$. Add in another $C/x$ term so that you can get the limiting value at $0$ to be exactly what you want. –  Hurkyl Mar 31 '12 at 11:20
I don't understand as to what do you mean by adding another C/x term. Can you please elaborate on this. Thank you. –  Shahab Mar 31 '12 at 11:26
I mean $F(x) = -2^{41/5}/(2x^2) + C/x + \langle \text{other stuff}\rangle$. If you're familiar with it, I'm just describing the first two terms of the Laurent series for $F(x)$. –  Hurkyl Mar 31 '12 at 12:21