Four integrals evaluated with Euler´s formula

Four integrals evaluated with Euler´s formula trick

In this post I intend to evaluate four basic integrals that will turn to be very useful in future posts for evaluating more complex and interesting integrals.

Usually these integrals are evaluated integrating by parts twice, which renders a quite tedious and long task. The idea here is to use Euler´s formula, namely:  e^{ix}= \cos(x) + i \sin(x)  to turn these integrals into simple exponential integrals which are very easy and straightforward to evaluate and consequently each pair of integrals will be evaluated through only one integral.

So lets get started!

The first pair of integrals that I want to consider are the Laplace transform of  sin(at)  and  cos(at).

I_{1}=\int_{0}^{\infty} e^{-s t} \cos (a t) d t(1)


I_{2}=\int_{0}^{\infty} e^{-s t} \sin (a t) d t(2)


I_{3}=\int_{0}^{\infty} e^{-s t} \cdot e^{-i a t} d t =\int_{0}^{\infty} e^{-(s+i a) t} d t(3)


I_{3}=-\left.\frac{e^{-(s+i a) t}}{s+i a}\right|_{0} ^{\infty}=\frac{1}{s+i a} \cdot \frac{s-i a}{s-i a}

And we get that:

\boxed{I_{3}=\int_{0}^{\infty} e^{-(s+i a) t} d t =\frac{s}{s^{2}+a^{2}}-i \frac{a}{s^{2}+a^{2}}}


Note that  I_{1}=\Re \left\{I_{3}\right\} and  I_{2}=\Im \left\{I_{3}\right\}  which gives us that:


\boxed{I_{1}=\int_{0}^{\infty} e^{-s t} \cos (a t) d t=\frac{s}{s^{2}+a^{2}}}


\boxed{I_{2}=\int_{0}^{\infty} e^{-s t} \sin (a t) d t=\frac{a}{s^{2}+a^{2}}}

The next pair of integrals looks very much like the previous one with except that now the limits of integration are from zero to one instead of zero to infinity and we have switched a with  2 \pi n  which leads to a different answer.

Consider the following:

J_{1}=\int_{0}^{1}e^{-tx}\cos(2 \pi n x) dx(4)


J_{2}=\int_{0}^{1}e^{-tx}\sin(2 \pi n x) dx(5)


J_{3}=\int_{0}^{1}e^{-tx}e^{-2 \pi i n x} dx = \int_{0}^{1}e^{-x(t+2 \pi i n )} dx(6)


J_{3}=-\frac{e^{-x(t+2 \pi i n )}}{t+2 \pi i n}\bigg|_{0}^{1}= \frac{1}{t+2 \pi i n}-\frac{e^{-(t+2 \pi i n )}}{t+2 \pi i n}


J_{3}=\frac{1}{t+2 \pi i n}-\frac{e^{-t}\Big( \cos(2 \pi n) - i \sin(2 \pi n ) \Big)}{t+2 \pi i n}


J_{3}=\frac{1-e^{-t}}{t+2 \pi i n} \cdot \frac{t-2 \pi i n}{t-2 \pi i n}


\boxed{J_{3}=\int_{0}^{1}e^{-x(t+ 2 \pi i n)}dx=\frac{t \cdot (1-e^{-t})}{t^2+(2 \pi n)^2}- i \cdot \frac{2 \pi n \cdot (1-e^{-t})}{t^2+(2 \pi n)^2} }


Note that   J_{1}= \Re\{J_{3}\}  and  J_{2}= \Im\{J_{3}\}  and we finally get the desired solutions:



\boxed{J_{1}=\int_{0}^{1}e^{-xt}\cos(2 \pi x)dx=\frac{t \cdot (1-e^{-t})}{t^2+(2 \pi n)^2} }




\boxed{J_{2}=\int_{0}^{1}e^{-xt}\sin( 2 \pi x )dx= \frac{2 \pi n \cdot (1-e^{-t})}{t^2+(2 \pi n)^2} }


Ricardo Albahari

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