∫_0^∞e^ stf(t)dt=Lf(t) ⇒∫_0^∞(sin tt)e^ stdt =L(sin tt) =∫_s^∞Lsin tds =∫_s^∞1s^2+1ds =[tan^ 1s]_s^∞ =π2 tan^ 1s Putting, s = 0 in the above equation ∫_0^∞sin t ...

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Fundamental Laws of Logarithms

Evaluate ∫0∞s...

Question

Evaluate $\int_{0}^{\infty} \frac{sin t}{t} d t$

π

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\frac{π}{2}

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\frac{π}{4}

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\frac{π}{8}

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Solution

The correct option is B $\frac{π}{2}$
$\int_{0}^{\infty} e^{- s t} f (t) d t = L$ {f(t)}
$\Rightarrow \int_{0}^{\infty} (\frac{sin t}{t}) e^{- s t} d t = L (\frac{sin t}{t})$
$= \int_{s}^{\infty} L$ { $sin t$ }ds
$= \int_{s}^{\infty} \frac{1}{s^{2} + 1} d s$
= $[{tan}^{- 1} s]_{s}^{\infty}$
$= \frac{π}{2} - {tan}^{- 1} s$
Putting, $s = 0$ in the above equation $\int_{0}^{\infty} \frac{sin t}{t} d t = \frac{π}{2} - {tan}^{- 1} 0$
$= \frac{π}{2}$

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Evaluate ∫∞0sinttdt

The correct option is B π2∫∞0e−stf(t)dt=L{f(t)} ⇒∫∞0(sintt)e−stdt=L(sintt) =∫∞sL{sint}ds =∫∞s1s2+1ds =[tan−1s]∞s =π2−tan−1s Putting, s=0 in the above equation ∫∞0sinttdt=π2−tan−10 =π2

Evaluate $\int_{0}^{\infty} \frac{sin t}{t} d t$