Question

The integral
\[ \frac{1}{\pi}\int_0^{\infty}\frac{x^{2026}}{(1+x^{2026})(1+x^2)}\,dx \] evaluates to (round off to two decimal places).

Hint:

Try the substitution x = 1/t to relate the integral to itself.

Answer 1

Correct Answer: 0.25

Step 1: Rewrite the integrand before doing anything else.
Notice that $\dfrac{x^{2026}}{1+x^{2026}}=1-\dfrac{1}{1+x^{2026}}$. So the integral becomes
$I=\displaystyle\int_0^{\infty}\left(1-\dfrac{1}{1+x^{2026}}\right)\dfrac{1}{1+x^2}\,dx=\int_0^{\infty}\dfrac{dx}{1+x^2}-\int_0^{\infty}\dfrac{dx}{(1+x^{2026})(1+x^2)}$.

Step 2: Evaluate the first piece.
$\displaystyle\int_0^{\infty}\dfrac{dx}{1+x^2}=\dfrac{\pi}{2}$, a standard arctangent result. Call the second piece $J$, so $I=\dfrac{\pi}{2}-J$.

Step 3: Show $J$ equals $I$ by substituting $x=1/t$ directly into $J$.
$J=\displaystyle\int_0^{\infty}\dfrac{dx}{(1+x^{2026})(1+x^2)}$. Put $x=1/t$, $dx=-dt/t^2$. Then $1+x^{2026}=\dfrac{t^{2026}+1}{t^{2026}}$ and $1+x^2=\dfrac{t^2+1}{t^2}$, so the integrand becomes
$\dfrac{t^{2026}}{t^{2026}+1}\cdot\dfrac{t^2}{t^2+1}\cdot\dfrac{1}{t^2}=\dfrac{t^{2026}}{(1+t^{2026})(1+t^2)}$.
This is exactly the original expression for $I$ with $t$ in place of $x$, so $J=I$.

Step 4: Combine the two facts.
We now have $I=\dfrac{\pi}{2}-J$ and $J=I$, so
$I=\dfrac{\pi}{2}-I\implies2I=\dfrac{\pi}{2}\implies I=\dfrac{\pi}{4}$.

Step 5: Divide by $\pi$.
$\dfrac{I}{\pi}=\dfrac{\pi/4}{\pi}=\dfrac{1}{4}=0.25$
\[\boxed{0.25}\]