The value of integral is: [2026]

Question

The value of integral $\int_{0}^{\infty} \frac{\log_{e} (x)}{x^{2} + 4} d x$ is: [2026]

Smart Achievers · Accepted Answer

(2)

Let $I = \int_{0}^{\infty} \frac{\log_{e} x}{x^{2} + 4} d x$

Put x = 2t $\Rightarrow$ dx = 2dt

$I = \int_{0}^{\infty} \frac{\log_{e} 2 t}{4 t^{2} + 4} \cdot 2 d t$

$= \frac{2}{4} \int_{0}^{\infty} \frac{\log_{e} 2 + \log_{e} t}{t^{2} + 1} d t$

$= \frac{1}{2} \int_{0}^{\infty} \frac{\log_{e} 2}{t^{2} + 1} d t + \frac{1}{2} \int_{0}^{\infty} \frac{\log_{e} t}{t^{2} + 1} d t$

$\Rightarrow I + I_{1} + I_{2}$

Now, $I_{1} = \frac{1}{2} \int_{0}^{\infty} \frac{\log_{e} 2}{t^{2} + 1} d t$

$= \frac{1}{2} \log_{e} 2 \cdot {(\tan^{- 1} t)}_{0}^{\infty}$

$= \frac{1}{2} \log_{e} 2 \cdot \frac{π}{4} = \frac{π}{4} \log_{e} 2$

$I_{2} = \frac{1}{2} \int_{0}^{\infty} \frac{\log_{e} t}{t^{2} + 1} d t$

Put, $t = \frac{1}{u} \Rightarrow d t = - \frac{1}{u^{2} d u}$

$= \frac{1}{2} \int_{\infty}^{0} \frac{- \log_{e} u}{\frac{1}{u^{2}} + 1} \cdot (\frac{- 1}{u^{2}}) d u$

$= - \frac{1}{2} \int_{0}^{\infty} \frac{\log_{e} u}{u^{2} + 1} d u \Rightarrow I_{2} = - I_{2} \Rightarrow I_{2} = 0$

$∴ I = \frac{π}{4} \log_{e} 2$

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