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∫ f(xⁿ+x⁻ⁿ)·lnx·dx/x from 0 to ∞ | JEE

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Question
∫0∞f ⁣(xn+x−n)ln⁡x dxx\displaystyle\int_{0}^{\infty}f\!\big(x^{n}+x^{-n}\big)\ln x\,\dfrac{dx}{x} equals
A00correct
Bf ⁣(xn+x−n)f\!\big(x^{n}+x^{-n}\big)
Ca function of xx
Dcannot be determined
Solution
Step 1: Substitute t=ln⁡xt=\ln x, so x=etx=e^{t} and dxx=dt\dfrac{dx}{x}=dt. The limits map as x→0+⇒t→−∞x\to 0^{+}\Rightarrow t\to-\infty and x→∞⇒t→∞x\to\infty\Rightarrow t\to\infty:
I=∫−∞∞f ⁣(ent+e−nt) t dt.I=\int_{-\infty}^{\infty}f\!\big(e^{nt}+e^{-nt}\big)\,t\,dt.
 Step 2: Examine the parity of the integrand. The argument ent+e−nt=2cosh⁡(nt)e^{nt}+e^{-nt}=2\cosh(nt) is an even function of tt, so f ⁣(ent+e−nt)f\!\big(e^{nt}+e^{-nt}\big) is even in tt. Step 3: Multiplying an even function by tt (odd) produces an odd integrand:
g(t)=f ⁣(ent+e−nt) t,g(−t)=−g(t).g(t)=f\!\big(e^{nt}+e^{-nt}\big)\,t,\qquad g(-t)=-g(t).
 Step 4: The integral of an odd function over the symmetric interval (−∞,∞)(-\infty,\infty) is zero:
I=∫−∞∞g(t) dt=0.I=\int_{-\infty}^{\infty}g(t)\,dt=0.
 Correct answer: (1)
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