Can you help me see why these integrals are the same?

[SamRoss]

I am reading "Inside Interesting Integrals" by Paul Nahin. Around pg. 59, he goes through a lengthy explanation of how to do the definite integral from 0 to infinity of ∫1/(x⁴+1)dx. However, he then simply writes down that this integral is equal to ∫x²/(x⁴+1)dx with the same limits. Now, it's easy to pop these into an integral calculator and see that they are in fact both equal to 1.1107... = (π√2)/4, but how can I see that they should be equal to each other without actually calculating the value for each? In other words, if you just showed me ∫1/(x⁴+1)dx = ∫x²/(x⁴+1)dx , I would be suspicious because of the different numerators but apparently I would be wrong. Why?

 

[Delta2]

Well these two are indeed interesting integrals as the book title says. You are right that at a first glance we wouldn't think that these two integrals are the same. But it turns out that they are the same, that's why they are interesting integrals!

My view is to brake the interval of integration to $[0,1]$ and $[1,+\infty)$.What happens is that the integral of $\frac{1}{x^4+1}$ in $[0,1]$ is greater than the integral of $\frac{x^2}{x^4+1}$ in the same interval. In the interval $[1,+\infty)$ the situation is reversed and in this interval $\frac{x^2}{x^4+1}$ is bigger but not as much as our intuition possible says, it is exactly much bigger as it is needed to cover the loss from the interval of $[0,1]$ so in the end both integrals come to a tie in the whole interval of $[0,+\infty)$.

 

[BvU]

Intriguing. Wolframalpha gives an analytical form:

and it takes (me) a while to notice the subtle difference.

Too big a challenge for me to uncover the path to come to such an expression...

 

[BvU]

Too big a challenge

Or rather: laziness . From looking at the answer it appears partial fraction decomposition is the way to go. Fancy tools they have nowadays !

 

[lurflurf]

$$\int_a^b=-\int_b^a$$
there is a theorem of elementary calculus called change of variable or substitution
see a good calculus book
$$\int_{g(a^+)}^{g(b^-)}f(x)dx=\int_{a^+}^{b^-}f(g(x))g^\prime(x)dx$$
f continuous on (a,b)
g differentiable with integrable derivative on (a,b)

let f be either of your functions
$$a=0$$
$$b=\infty$$
$$g(x)=1/x$$
this will establish the equality

 

[alan2]

Depends on what you mean by the same. They happen to evaluate to the same value when the upper limit is infinite but they are not the same integral. Plug in any finite upper bound and you will find they are different integrals. In fact, plot the two integrands and you’ll find they are quite different.

 

[lurflurf]

^
The claim was
$$\int_a^\infty \frac{dx}{1+x^4}=\int_0^{1/a} \frac{x^2 dx}{1+x^4}$$
when
$$a=0^+$$
but fell free to take for a anything in $$(0,\infty)$$

 

[alan2]

^
The claim was
$$\int_a^\infty \frac{dx}{1+x^4}=\int_0^{1/a} \frac{x^2 dx}{1+x^4}$$
when
$$a=0^+$$
but fell free to take for a anything in $$(0,\infty)$$

I could have been more clear. You gave the correct substitution above but notice that another poster plugged the indefinite integrals into a solver. Those are very different although I assume you could take limits on those expressions, I’m too lazy to check. I’m not sure the OP understands yet what is going on.

 

[lurflurf]

^
we have for $$a\in(0,\infty)$$
$$\int_a^{1/a} \frac{x^2-1}{x^4+1}dx=0$$
this is due to symmetry that is easy to miss
we can make the symmetry more obvious through the Euler-Weierstrass substitution x=tan(t/2)
$$\int_0^\infty \frac{1}{x^4+1}dx=\int_0^\pi \frac{1+\cos(t)}{3+\cos(2t)}dt$$
$$\int_0^\infty \frac{x}{x^4+1}dx=\int_0^\pi \frac{\sin(t)}{3+\cos(2t)}dt$$
$$\int_0^\infty \frac{x^2}{x^4+1}dx=\int_0^\pi \frac{1-\cos(t)}{3+\cos(2t)}dt$$
the first and third are reflections of each other about $$t=\pi/2$$ which is easier to recognize
nice when versine and versine show up