Integral by Euler’s Formula

We compute the following integral

(1)   \begin{align*} \int_{1}^{\infty}{dx \over x\, (x^2 + 1)} \end{align*}

Performing a partial fraction expansion of the integrand

(2)   \begin{align*} {1 \over x\, (x^2 + 1)} \,=\, {1 \over x\, (x + i)(x - i)} \,=\, {1 \over x} - {1 \over 2(x + i)} - {1 \over 2(x - i)} \end{align*}

Thus we have

(3)   \begin{align*} \int_{1}^{\infty}dx\, \bigg( {1 \over x} - {1 \over 2(x + i)} - {1 \over 2(x - i)} \bigg) \,&=\, \Big(\ln x - 2\ln(x + i) - 2\ln(x - i)\Big)\Big|_{1}^{\infty} \\[0.5em] \,&=\, \bigg(\ln{x \over \sqrt{x^2 + 1}}\bigg)\bigg|_{1}^{\infty} \\[1.1em] \,&=\, \ln\sqrt{2} \end{align*}

Let us see another integral

(4)   \begin{align*} \int_{0}^{\infty}dx\, \sin(\alpha x)\, e^{-\beta x} \end{align*}

Using Euler’s Formula

(5)   \begin{align*} \sin(\alpha x) \,=\, {e^{i\alpha x} - e^{-i\alpha x} \over 2i} \end{align*}

we have

(6)   \begin{align*} \int_{0}^{\infty}dx\, \sin(\alpha x)\, e^{-\beta x} \,&=\, {1 \over 2i}\int_{0}^{\infty}dx\, \Big(e^{(i\alpha -\beta) x} - e^{-(i\alpha + \beta) x}\Big) \\ \,&=\, {1 \over 2i}\bigg({e^{(i\alpha -\beta)x} \over i\alpha -\beta} - {e^{-(i\alpha + \beta) x} \over -(i\alpha + \beta)}\Big)\bigg|_{0}^{\infty} \\ \,&=\, {1 \over 2i}\bigg({1 \over \beta - i\alpha} - {1 \over \beta + i\alpha}\Big) \\ \,&=\, {\alpha \over \alpha^2 + \beta^2} \end{align*}

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