If a- b 0 and -0-lnbxx2-a2 dx-2a- then the value of ab is

Let I=∫_0^∞ln(bx)x^2+a^2 dx Assuming x = a tanθ ⇒ dx=a ^2θ dθ Now, nbsp; I=∫_0^π/2ln(ab tanθ)a^2 θ dθa^2^2 θ ⇒ I = 1a∫_0^π/2(ln ab+lntanθ) dθ ⇒ I=π2aln ab+∫_ ...

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Standard XIII

Mathematics

Property 6

If a, b 0 and...

Question

If $a, b > 0$ and $\infty \int 0 \frac{ln (b x)}{x^{2} + a^{2}} d x = \frac{π}{2 a},$ then the value of $a b$ is

1

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\frac{1}{e}

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e

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\frac{1}{2}

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Solution

The correct option is C $e$
Let $I = \infty \int 0 \frac{ln (b x)}{x^{2} + a^{2}} d x$
Assuming $x = a tan θ$
$\Rightarrow d x = a {sec}^{2} θ d θ$

Now, $I = π / 2 \int 0 \frac{ln (a b tan θ) a {sec}^{2} θ d θ}{a^{2} {sec}^{2} θ}$
$\Rightarrow I = \frac{1}{a} π / 2 \int 0 (ln a b + ln tan θ) d θ \Rightarrow I = \frac{π}{2 a} ln a b + π / 2 \int 0 ln tan θ d θ$
Replacing $θ \to \frac{π}{2} - θ$ , we get
$I = \frac{π}{2 a} ln a b + π / 2 \int 0 ln cot θ d θ$
Adding both, we get
$2 I = \frac{π}{a} ln a b + π / 2 \int 0 [ln tan θ + ln cot θ] d θ \Rightarrow 2 I = \frac{π}{a} ln a b + π / 2 \int 0 [ln 1] d θ \Rightarrow \frac{π}{2 a} = \frac{π}{2 a} ln a b \Rightarrow ln (a b) = 1 ∴ a b = e$

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If a,b>0 and ∞∫0ln(bx)x2+a2 dx=π2a, then the value of ab is

If $a, b > 0$ and $\infty \int 0 \frac{ln (b x)}{x^{2} + a^{2}} d x = \frac{π}{2 a},$ then the value of $a b$ is