Proof of some Trig identities through Euler´s formula and evaluation of two integrals Involving LogGamma function

Proof of some Trig identities through Euler´s formula and evaluation of two integrals Involving LogGamma function

Today I want to show how some trigonometric identities can be proved from Euler´s formula and then compute two Integrals involving the Loggamma benefiting from two of these identities.

1.Magic through Euler´s formula

Euler´s formula is given by

$e^{ix}=\cos x + i \sin x$

From it, we can easily derive some useful trigonometric identities.

Consider

$e^{ix}e^{iay}=e^{ix+iay}=e^{i(x+ay)}=\cos( x+ay) + i \sin( x+ay)$ (1.1)

On the other hand we have

$e^{ix}e^{iay}=[\cos x + i \sin x]\cdot[\cos ay + i \sin ay]$

$=\cos x \cos ay+ i \cos x \sin ay + i sin x \cos ay- \sin x \sin ay$

$= \cos x \cos ay- \sin x \sin ay+ i \big(\cos x \sin ay+ sin x \cos ay \big)$ (1.2)

Equating Real and Imaginary parts of (1.1) and (1.2) we get

$\boxed{\cos( x+ay)=\cos x \cos ay- \sin x \sin ay}$ (1.3)

$\boxed{\sin( x+ay)=\cos x \sin ay+ sin x \cos ay }$ (1.4)

The difference formulas can be derived in the same fashion

$e^{ix}e^{-iay}=e^{i(x-ay)}=\cos( x-ay) + i \sin( x-ay)$

after the algebra we get

$\boxed{\cos( x-ay)=\cos x \cos ay+ \sin x \sin ay}$ (1.5)

$\boxed{\sin( x-ay)=sin x \cos ay-\cos x \sin ay}$ (1.6)

Double angle formula

If we consider

$(e^{ix})^2=e^{2ix}= \cos(2x)+i \sin(2x)$ (2.1)

$(e^{ix})^2=[\cos(x)+i \sin(x)]^2$

$=\cos^2(x)+ 2i \cos(x) \sin (x) - sin^2(x)$

$e^{2ix}=\cos^2(x)- sin^2(x)+2i \cos(x) \sin (x)$ (2.2)

Equating Real and Imaginary parts of (2.1) and (2.2) we get

$\boxed{\cos(2x)=\cos^2(x)- sin^2(x)}$ (2.3)

$\boxed{\sin(2x)=2\cos(x) \sin (x)}$ (2.4)

From the identity

$\cos^2 x+ \sin^2 x = 1$ (2.5)

$\cos^2 x=1-\sin^2 x$

we can re wright (2.3) as

$\boxed{\cos(2x)=1- 2sin^2(x)}$ (2.6)

or

$\boxed{sin^2(x)=\frac{1-\cos(2x)}{2}}$ (2.7)

From (2.5) we get that

$\sin^2 x=1-\cos^2 x$

substituting it in (2.3) we get

$\cos(2x)=\cos^2(x)- (1-\cos^2 x)$

$\boxed{\cos^2 x=\frac{1+\cos(2x)}{2}}$ (2.8)

Evaluation of two integrals

Based on this formulas lets solve two scary looking integrals.

$I_{1}=\int_{0}^{1}\log(\Gamma(x))\cos^2(\pi x) dx$

and

$I_{2}=\int_{0}^{1}\log(\Gamma(x))\sin^2(\pi x) dx$

Based on (2.8) we can re-wright $I_{1}$ as

$I_{1}=\int_{0}^{1}\log(\Gamma(x)) \Big(\frac{1+\cos(2 \pi x)}{2} \Big) dx$

$=\frac{1}{2}\int_{0}^{1}\log(\Gamma(x))dx+\int_{0}^{1}\log(\Gamma(x)\cos(2 \pi x))dx$ (3.1)

From this post we know that

(3.2)

and that $\int_{0}^{1}\log(\Gamma(x))dx=\frac{\log(2 \pi)}{2}$

and

$\int_{0}^{1}\log(\Gamma(x)\cos(2 \pi k x))dx=\frac{1}{4k}$ (3.3)

setting k=1 in (3.3) and substituting the results in (3.1) we get that

$\boxed{\int_{0}^{1}\log(\Gamma(x))\cos^2(\pi x) dx= \frac{\log(2 \pi)}{4}+\frac{1}{8}}$

$I_{2}$ is computed in a similar way

$I_{2}=\int_{0}^{1}\log(\Gamma(x))\sin^2(\pi x) dx$

Substituting (2.7) in the above integral we get

$I_{1}=\int_{0}^{1}\log(\Gamma(x)) \Big(\frac{1-\cos(2 \pi x)}{2} \Big) dx$

splitting into two integrals and following the same steps as in the previous integral we get that

$\boxed{I_{2}=\int_{0}^{1}\log(\Gamma(x))\sin^2(\pi x) dx=\frac{\log(2 \pi)}{4}-\frac{1}{8}}$

Ricardo Albahari

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