GATE 2026 EE Q.65: solution and answer

Official answer: 0.25Source: Official answer key, GATE 2026 EE (IIT Guwahati)

1. Let $I={\int }_0^{\infty }\frac{x^{2026}}{(1+x^{2026})(1+x^2)}dx$. For $x>0$ the factor $\frac{x^{2026}}{1+x^{2026}}$ lies between $0$ and $1$, so the integrand is at most $\frac{1}{1+x^2}$ and the integral converges. There is no usable antiderivative, so use the reciprocal substitution $x=\frac{1}{t}$, $dx=-\frac{dt}{t^2}$. The limits $0\to \infty$ become $\infty \to 0$, and the minus sign flips them back to $0\to \infty$.

2. Substitute into the integrand: $$\frac{t^{-2026}}{(1+t^{-2026})(1+t^{-2})}\cdot \frac{1}{t^2}=\frac{1}{1+t^{2026}}\cdot \frac{t^2}{1+t^2}\cdot \frac{1}{t^2}=\frac{1}{(1+t^{2026})(1+t^2)}$$ (multiply the numerator and denominator of the first fraction by $t^{2026}$, and of the second by $t^2$).

3. Renaming $t$ back to $x$, the same integral can therefore also be written as $I={\int }_0^{\infty }\frac{1}{(1+x^{2026})(1+x^2)}dx$.

4. Add the two forms of $I$. The factor $1+x^{2026}$ cancels: $$2I={\int }_0^{\infty }\frac{x^{2026}+1}{(1+x^{2026})(1+x^2)}dx={\int }_0^{\infty }\frac{dx}{1+x^2}=[{\tan }^{-1}x{]}_0^{\infty }=\frac{\pi }{2}$$

5. So $I=\frac{\pi }{4}$, and the required value is $\frac{1}{\pi }I=\frac{1}{4}=0.25$.

6. Check: the exponent 2026 never entered the result. The same argument gives $\pi /4$ for any real power $n$ in place of 2026 (for $n=0$ the integrand is simply $\frac{1}{2}\cdot \frac{1}{1+x^2}$, which integrates to $\pi /4$, as it should).