Question

${\int }_0^{\infty }f(x+\frac{1}{x}).\frac{\ln x}{x}dx$

• A. Is equal to zero
• B. Is equal to one
• C. Is equal to $\frac{1}{2}$
• D. Can not be evaluated

Answer 1

Answer: (A)

Is equal to zero

Let: $lnx=t\Rightarrow x=e^t$

$\Rightarrow \frac{1}{x}dx=dt$

As "x" varies from $0$ to $\infty$ "$lnx$ $\left[t\right]$" varies $-\infty$ to $\infty$.

Now,

${\int }_0^{\infty }f(x+\frac{1}{x}).\frac{lnx}{x}dx$

$\Rightarrow {\int }_{-\infty }^{\infty }f(e^t+e^{-t}).tdt=F\left(t\right)$

Now,

Using properties of definite integral:

Here we can see above function is an odd function i.e $F(-t)=-F\left(t\right)$

therefore on integrating from $-\infty$ to $\infty$ sum of area of $odd$ $function$ is $zero.$

$\Rightarrow {\int }_{-\infty }^{\infty }f(e^t+e^{-t}).tdt=0$

Thus,

${\int }_0^{\infty }f(x+\frac{1}{x}).\frac{lnx}{x}dx=0$

Hence, correct option is $"A"$