If A=∫0∞sinxxdx, then the value of the integral ∫0∞sinaxcosbxxdx, is equal to

A=∫0∞sinxxdx I=∫0∞sinaxcosbxxdx=12[∫0∞sin(a+b)x+sin(a−b)xxdx] I=12[∫0∞sin(a+b)xxdx+∫0∞sin(a−b)xxdx] I1=∫0∞sin(a−b)xxdxLet (a+b)x=t⇒I1=∫0(a+b)(→∞)sinttdx I2=∫0∞sin(a−b)xxdxLet (a+b)x=t⇒I2=∫0(a+b)(→∞)sinttdtI=12(I1+I2) If a+b>0; then I1=A. If a−b>0, then I2=AIf a−b b>0⇒I=12(A+A)=A b>a>0⇒I=12(A−A)=0

If A=∫0∞sinxxdx, then the value of the integral ∫0∞sinaxcosbxxdx, is equal to

A=∫0∞sinxxdx I=∫0∞sinaxcosbxxdx=12[∫0∞sin(a+b)x+sin(a−b)xxdx] I=12[∫0∞sin(a+b)xxdx+∫0∞sin(a−b)xxdx] I1=∫0∞sin(a−b)xxdxLet (a+b)x=t⇒I1=∫0(a+b)(→∞)sinttdx I2=∫0∞sin(a−b)xxdxLet (a+b)x=t⇒I2=∫0(a+b)(→∞)sinttdtI=12(I1+I2) If a+b>0; then I1=A. If a−b>0, then I2=AIf a−b b>0⇒I=12(A+A)=A b>a>0⇒I=12(A−A)=0

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If $A = \int_{0}^{\infty} \frac{\sin x}{x} d x,$ then the value of the integral $\int_{0}^{\infty} \frac{\sin a x \cos b x}{x} d x,$ is equal to

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$A if a > b > 0$

$2 A if a + b = 0$

$\frac{A}{a + b}$

$0 if b > a > 0$

answer is A, D.

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Detailed Solution

$A = \int_{0}^{\infty} \frac{\sin x}{x} d x$
$I = \int_{0}^{\infty} \frac{\sin a x \cos b x}{x} d x = \frac{1}{2} [\int_{0}^{\infty} \frac{\sin (a + b) x + \sin (a - b) x}{x} d x]$
$I = \frac{1}{2} [\int_{0}^{\infty} \frac{\sin (a + b) x}{x} d x + \int_{0}^{\infty} \frac{\sin (a - b) x}{x} d x]$
$I_{1} = \int_{0}^{\infty} \frac{\sin (a - b) x}{x} d x$
Let $(a + b) x = t \Rightarrow I_{1} = \int_{0}^{(a + b) (\to \infty)} \frac{\sin t}{t} d x$
$I_{2} = \int_{0}^{\infty} \frac{\sin (a - b) x}{x} d x$
Let $(a + b) x = t \Rightarrow I_{2} = \int_{0}^{(a + b) (\to \infty)} \frac{\sin t}{t} d t$
$I = \frac{1}{2} (I_{1} + I_{2})$
If $a + b > 0;$ then $I_{1} = A .$ If $a - b > 0,$ then $I_{2} = A$
If $a - b < 0;$ then $I_{2} = \int_{0}^{- \infty} \frac{\sin t}{t} d t = \int_{0}^{\infty} \frac{\sin t}{t} d t = A$
Now, $a > b > 0 \Rightarrow I = \frac{1}{2} (A + A) = A$
$b > a > 0 \Rightarrow I = \frac{1}{2} (A - A) = 0$