# Calculating the integral with an undefined function f(x)

I am having a problem with an excersice of calculating the integral

I= $$\int_{0}^{a}\frac{1}{1+f(x)}\,dx$$

and we know that $$f(x)f(a-x)=1$$ , $$a>0$$ and $$f(x)$$ is continous and positive on the interval $$[0,a]$$

i've tried manipulating the expression with the fact that $$f(x)= \ \frac{1}{f(a-x)}$$ in order to find something to cancel or something that makes me solve the integral "nicely"

However after 20 minutes i have no idea what to do, or what i am missing

so if anyone could give me a tip or point me in the right direction i would appreciate it

$$\int_0^a\frac{1}{1+f(x)}\,dx = \int_0^a\frac{1}{1+f(a-x)}\,dx = \int_0^a\frac{1}{1+\frac 1{f(x)}}\,dx = \int_0^a\frac{f(x)}{1+f(x)}\,dx,$$ where in the first transformation we changed variables $$x = a-y$$. Hence, $$2I = I + I = \int_0^a\frac{1}{1+f(x)}\,dx + \int_0^a\frac{f(x)}{1+f(x)}\,dx = \int_0^a1\,dx = a.$$ Thus, $$I = \frac a2$$.

• I don't understand why $\int_0^a\frac{1}{1+f(x)}\,dx = \int_0^a\frac{1}{1+f(a-x)}\,dx$ because we don't got $f(x)=f(a-x)$ we got $f(x)=1/f(a-x)$. Help? Mar 5, 2021 at 23:32
• @PrimeMover Change of variables. I edited my answer. Mar 5, 2021 at 23:33
• I'm still a bit slow, wouldn't that make the second integral negative? $dx = -dy$ and all that? Mar 5, 2021 at 23:36
• @PrimeMover No. Watch the integral boundaries. $-\int_a^0 = \int_0^a$ Mar 5, 2021 at 23:40
• Oh right okay no worries Mar 5, 2021 at 23:42

If the answer is independent of $$f(x)$$, then you can choose one consistent with the stated conditions, e.g., $$f(x)=1$$, in which case we find:

$$\int\limits_{x=0}^a \frac{1}{1 + 1} dx = \frac{a}{2}.$$

• I upvoted this answer, but we need to be careful. The question does not actually say that the answer is independent of $f$. Mar 6, 2021 at 8:53
• @BrianDrake: True enough, but one can say that $a/2$ is one valid solution! Mar 6, 2021 at 18:15

Same answer, different approach. By symmetry, \begin{align} \int_0^a\frac1{1+f(x)}\,\mathrm{d}x &=\frac12\left(\int_0^a\frac1{1+f(x)}\,\mathrm{d}x+\int_0^a\frac1{1+f(a-x)}\,\mathrm{d}x\right)\tag1\\ &=\frac12\int_0^a\left(\frac1{1+f(x)}+\frac1{1+f(a-x)}\right)\,\mathrm{d}x\tag2\\ &=\frac12\int_0^a\frac{\color{#C00}{1}+f(x)+\color{#C00}{1}+f(a-x)}{\color{#090}{1}+f(x)+f(a-x)+\color{#090}{f(x)f(a-x)}}\,\mathrm{d}x\tag3\\ &=\frac12\int_0^a\frac{\color{#C00}{2}+f(x)+f(a-x)}{\color{#090}{2}+f(x)+f(a-x)}\,\mathrm{d}x\tag4\\[3pt] &=\frac a2\tag5 \end{align} Explanation:
$$(1)$$: by substituting $$x\mapsto a-x$$, we get $$\int_0^a\frac1{1+f(x)}\,\mathrm{d}x=\int_0^a\frac1{1+f(a-x)}\,\mathrm{d}x$$
$$\phantom{\text{(1):}}$$ average the two
$$(2)$$: sum of integrals is the integral of the sum
$$(3)$$: $$\frac1u+\frac1v=\frac{u+v}{uv}$$
$$(4)$$: the sum of the integrands is $$1$$ since $$f(x)f(a-x)=1$$
$$(5)$$: integrate

• I don't understand why you are adding, or why did you separate the original integral into to with the second one being the same but with $$f(a-x)$$ would'nt that just change the value? Mar 5, 2021 at 23:47
• Again: Change of variables. Or as robjohn likes to say: "By symmetry". ;-) Mar 5, 2021 at 23:56
• @Ramirogenta: substituting $x\mapsto a-x$, if done consistently, leaves the integral unchanged. If you prefer, you can add a step by substituting $x=a-y$ then another substituting $y=x$.
– robjohn
Mar 6, 2021 at 0:04
• You wrote “the sum of the integrands is $1$”, which is the result of equation $(3)$. Perhaps it would be better to explain the source of equation $(3)$, like you did for the other equations. In this case, it involves $f(x)f(a-x)=1$. Mar 6, 2021 at 8:58
• @BrianDrake: I'm sorry, I have added a step that I had previously thought unnecessary. I guess I was wrong.
– robjohn
Mar 6, 2021 at 10:49