integral with two parameters

**Random Variable** · July 3rd 2011, 11:50 PM

$\int_{0}^{1} \frac{\sin(p \ln x) \cos(q \ln x)}{\ln x} \ dx$ where $p, q >0$

To me this integral is just begging to be written as a triple integral. But I can't think of a way of doing it that will make evaluating the integral any easier.

**chisigma** · July 4th 2011, 12:46 AM

Originally Posted by Random Variable

$\int_{0}^{1} \frac{\sin(p \ln x) \cos(q \ln x)}{\ln x} \ dx$ where $p, q >0$

To me this integral is just begging to be written as a triple integral. But I can't think of a way of doing it that will make evaluating the integral any easier.

Setting $\ln x = t$ the integral becomes...

$I= \int_{0}^{\infty} \frac{\sin p t \ \cos q t }{t}\ e^{-t}\ dt = \mathcal{L} \{\frac{\sin p t \ \cos q t }{t}\}_{s=1}$ (1)

Now You have to remember that...

$\sin p t\ \cos q t = \frac{1}{2}\ \{\sin (p-q) t + \sin (p+q) t\}$ (2)

... and that...

$\mathcal{L} \{\frac{\sin a t}{t}\} = \int_{s}^{\infty} \frac{a}{a^{2}+ u^{2}}\ du = \tan^{-1} \frac{a}{s}$ (3)

... and, combining (1), (2) and (3) You obtain...

$I= \frac{1}{2}\ \{ \tan^{-1} (p-q) + \tan^{-1} (p+q) \}$ (4)

Kind regards

$\chi$ $\sigma$

**Random Variable** · July 4th 2011, 04:54 AM

I'm pretty sure you meant $t = - \ln x$ .

Interestingly, Maple returns the wrong answers for different values of the parameters while Wolfram Alpha returns the correct values. Like for p = 3 and q =2, Maple returns a negative value of $- \frac{1}{2} \arctan \Big( \frac{3}{2} \Big)$ which is clearly absurd when you look at the graph.

**Danny** · July 4th 2011, 05:58 AM

Here's another way. Define

$I(p,q) = \int_0^1 \frac{\sin p \ln x \cos q \ln x}{\ln x} dx$

noting that $I(0,q) = 0$ . Then make the substitution $u = \ln x$ (as suggested) so

$I = \int_{- \infty}^0 \frac{\sin p u \cos q u \;e^u}{u} du.$

Then

$I_p = \int_{- \infty}^0 \cos p u \cos q u e^u du$ .

Now let $\cos pu \cos qu = \frac{1}{2} \cos (p-q)u + \frac{1}{2}\cos (p+q)u$ .

Integration by part twice gives

$I_p = \frac{1}{2} \frac{1}{1 + (p-q)^2 } + \frac{1}{2} \frac{1}{1 + (p+q)^2}$

Integrating gives

$I = \frac{1}{2} \tan^{-1} (p-q) + \frac{1}{2} \tan^{-1} (p+q) + F(q)$

Since $I(0,q) = 0$ then $F(q) = 0$ thus giving what chisigma gave.

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