MHB Integrals: Show $\alpha$ Not Int Multiple of $\pi$ and $s, \lambda >0$

[polygamma]

1) Show that for $\alpha$ not an integer multiple of $\pi$, $\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \Big( \frac{\cos x + \sin x}{\cos x - \sin x} \Big)^{\cos \alpha} \ dx = \frac{\pi}{2 \sin \left(\pi \cos^{2} \frac{\alpha}{2} \right)} $.2) Show that for $s,\lambda >0$ and $0 \le \alpha < \frac{\pi}{2}$, $\displaystyle \int_{0}^{\infty} x^{s-1} e^{-\lambda x \cos \alpha} \cos(\lambda x \sin \alpha) \ dx = \frac{\Gamma(s) \cos (\alpha s)}{\lambda^{s}}$.

 

[alyafey22]

I think there is a mistake in the second question $$t^{s-1}$$ is completely independent of the integral so it can be taken out of the integration .

Maybe , you meant instead $$x^{s-1}$$

 

[polygamma]

I think there is a mistake in the second question $$t^{s-1}$$ is completely independent of the integral so it can be taken out of the integration .

Maybe , you meant instead $$x^{s-1}$$

Yes. That's what I meant. Thanks.

 

[alyafey22]

1) Show that for $\alpha$ not an integer multiple of $\pi$, $\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \Big( \frac{\cos x + \sin x}{\cos x - \sin x} \Big)^{\cos \alpha} \ dx = \frac{\pi}{2 \sin \left(\pi \cos^{2} \frac{\alpha}{2} \right)} $.

Cool problem : $$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \Big( \frac{\cos x + \sin x}{\cos x - \sin x} \Big)^{\cos \alpha} \ dx$$

$$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{1+\sin(2x)}{\cos(2x)} \right) ^{\cos \alpha} \, dx$$

Now we use the change of variable : $$t=\frac{\pi}{4}-x$$

$$\int^{\frac{\pi}{2}}_0 \left(\frac{1+\cos 2t }{\sin 2t } \right)^{\cos \alpha }\, dt$$

$$ \int ^{\frac{\pi}{2}}_0 (\cos t )^{\cos \alpha } \cdot (\sin t)^{-\cos \alpha }dt$$

Now using the beta identity we get :

$$\frac{1}{2} \text{B} \left( \frac{1+\cos \alpha }{2}, \frac{1-\cos \alpha }{2}\right) \, = \frac{1}{2} \Gamma \left(\frac{1+\cos \alpha }{2}\right) \cdot \Gamma \left(\frac{1-\cos \alpha }{2}\right)$$

Now we use the reflection formula to get :

$$\frac{\pi}{2 \sin \left( \pi \cos^2 \left( \frac{\alpha}{2} \right) \right) }$$

 

[polygamma]

A slightly different approach is $ \displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \Big( \frac{\cos x + \sin x}{\cos x - \sin x} \Big)^{\cos \alpha} \ dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \Bigg( \frac{ \sin \left(\frac{\pi}{4} +x \right)}{\sin \left(\frac{\pi}{4}-x\right)}\Bigg)^{\cos \alpha} dx $

$ \displaystyle = \int_{0}^{\frac{\pi}{2}} \Bigg( \frac{ \sin \left(\frac{\pi}{2} - u \right)}{\sin u}\Bigg)^{\cos \alpha} du = \int_{0}^{\frac{\pi}{2}} (\cos u)^{\cos \alpha} (\sin u)^{-\cos \alpha} \ du$

 

[alyafey22]

I solved the second one , should I post the solution ?

 

[alyafey22]

2) Show that for $s,\lambda >0$ and $0 \le \alpha < \frac{\pi}{2}$, $\displaystyle \int_{0}^{\infty} x^{s-1} e^{-\lambda x \cos \alpha} \cos(\lambda x \sin \alpha) \ dx = \frac{\Gamma(s) \cos (\alpha s)}{\lambda^{s}}$.

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Here is the solution for the second one :

Spoiler

$$\cos(\lambda x \sin \alpha) = \frac{e^{ i\lambda x \sin \alpha}+e^{-i\lambda x \sin \alpha}}{2}$$

$$\int_{0}^{\infty} x^{s-1} e^{-(\lambda x \cos \alpha )} \left( \frac{e^{ i\lambda x \sin \alpha}+e^{-i\lambda x \sin \alpha}}{2} \right) dx $$

$$\frac{1}{2}\int_{0}^{\infty} x^{s-1} e^{-\lambda x (\cos \alpha -i\sin \alpha)}\, dx+\, \frac{1}{2}\int_{0}^{\infty} x^{s-1} e^{-\lambda x (\cos \alpha +i\sin \alpha)} \, dx$$

$$\frac{1}{2}\int_{0}^{\infty} x^{s-1} e^{- \left(\lambda \,e^{i \alpha}\,x\right)}\, dx+\, \frac{1}{2}\int_{0}^{\infty} x^{s-1} e^{- \left(\lambda \,e^{-i \alpha}\, x\right)} \, dx$$

Then we use the Laplace transform :

$$\frac{\Gamma(s)}{ 2\left(\lambda \,e^{i \alpha}\right)^s }\,+\,\frac{\Gamma(s)}{2\left(\lambda \,e^{-i \alpha}\right)^s }\,=\frac{\Gamma(s)}{\lambda^s} \left( \frac{\,e^{i \alpha \, s}+\, e^{-i \alpha \, s}}{2} \right)\, = \, \frac{\Gamma(s) \cos (\alpha s)}{\lambda^{s}}$$

 

[alyafey22]

If you have another approach, please share it ...

 

[polygamma]

Here's an alternative solution using contour integration.

Let $f(z) = z^{s-1}e^{-\lambda z}$ and integrate around a sector in the first quadrant that forms an angle of $\alpha$ with the positive real axis. The contour needs to be indented at the origin since $z=0$ is a branch point.

The integral evaluates to zero along the big arc and the small arc in the limit. Just use Jordan's inequality.

So we have $\displaystyle \int_{0}^{\infty} f(x) dx - \int_{0}^{\infty} f(te^{ia}) \ dt =0 $

or $ \displaystyle \int_{0}^{\infty} x^{s-1} e^{-\lambda x} \ dx - e^{ias} \int_{0}^{\infty} t^{s-1} e^{-\lambda t \cos \alpha} e^{-i \lambda t \sin a} \ dt =0$

$\displaystyle \implies \int_{0}^{\infty} t^{s-1}e^{-\lambda t \cos \alpha} \Big(\cos(\lambda t \sin \alpha)- i \sin (\lambda t \sin\alpha) \Big) \ dt = e^{-ias} \int_{0}^{\infty}x^{s-1} e^{-\lambda x} \ dx = e^{-ias} \frac{\Gamma(s)}{\lambda^{s}}$

$ \displaystyle = \Big(\cos (as) - i \sin (as) \Big) \frac{\Gamma(s)}{\lambda^{s}}$

Now just equate the real parts on both sides of the equation.