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: Name the given integral.
Let
\[ I=\int_0^{\infty}\frac{x^{2026}}{(1+x^{2026})(1+x^2)}\,dx \]
so that the quantity we want is \(\dfrac{I}{\pi}\).

Step 2: Define a companion integral.
Let
\[ J=\int_0^{\infty}\frac{1}{(1+x^{2026})(1+x^2)}\,dx \]

Step 3: Add \(I\) and \(J\).
\[ I+J=\int_0^{\infty}\frac{x^{2026}+1}{(1+x^{2026})(1+x^2)}\,dx=\int_0^{\infty}\frac{1}{1+x^2}\,dx \]
This last integral is a standard one:
\[ \int_0^{\infty}\frac{dx}{1+x^2}=\Big[\tan^{-1}x\Big]_0^{\infty}=\frac{\pi}{2} \]
So
\[ I+J=\frac{\pi}{2} \]

Step 4: Show \(I=J\) using the substitution \(x=\dfrac{1}{t}\).
In \(I\), put \(x=\dfrac{1}{t}\), so \(dx=-\dfrac{dt}{t^2}\), and as \(x\) runs from \(0\) to \(\infty\), \(t\) runs from \(\infty\) to \(0\).
\[ 1+x^{2026}=1+\frac{1}{t^{2026}}=\frac{t^{2026}+1}{t^{2026}},\qquad 1+x^2=\frac{t^2+1}{t^2} \]
So
\[ I=\int_0^{\infty}\frac{t^{-2026}}{\left(\frac{t^{2026}+1}{t^{2026}}\right)\left(\frac{t^2+1}{t^2}\right)}\cdot\frac{dt}{t^2} \]
The powers of \(t\) simplify as
\[ t^{-2026}\times t^{2026}\times t^2\times\frac{1}{t^2}=1 \]
leaving
\[ I=\int_0^{\infty}\frac{dt}{(1+t^{2026})(1+t^2)}=J \]

Step 5: Solve for \(I\).
Since \(I=J\) and \(I+J=\dfrac{\pi}{2}\),
\[ 2I=\frac{\pi}{2}\implies I=\frac{\pi}{4} \]

Step 6: Divide by \(\pi\) as asked.
\[ \frac{I}{\pi}=\frac{\pi/4}{\pi}=\frac{1}{4}=0.25 \]

Final Answer:
\[ \boxed{0.25} \]