# Using "convergence factor" to transform integral.

A method to evaluate the integral $$\int^\infty_{-\infty}\frac{\sin{x}}{x}dx$$ is using the fact that $$\int^\infty_0e^{-xt}dt=\frac{1}{x} \tag1$$ then \begin{align} \int^\infty_{-\infty}\frac{\sin{x}}{x}dx &=2\int^\infty_{0}\frac{\sin{x}}{x}dx\\ &=2\int^\infty_0\int^\infty_0e^{-xt}\sin{x}dxdt\\ &=2\int^\infty_0\frac{1}{1+t^2}dt\\ &=\pi \end{align}

In some Chinese books, "$e^{-xt}$" is regarded as the "convergence factor( 收敛因子 in Chinese)" of "$\frac{1}{x}$", even though I don't know what exactly the term is in English, I wonder if there is any other "convergence factor" and integral like "$e^{-xt}$" → $\frac{1}{x}$ →"$\int^\infty_{-\infty}\frac{\sin{x}}{x}dx$"?

• Are you asking in hopes of evaluating $\int_{-\infty}^\infty \frac{\sin(x)}{x}dx$ in a new way, or do you want a reference for the technique underlying it? Commented Aug 30, 2017 at 6:54
• The term "convergence factor" looks convenient. An important fact is that, by using this factor, you introduce a new parameter $t$ and in this case, it is to be related to the Laplace transform. If we focus on the idea of functional transforms (as is Laplace transform), and to answer your question this "convergence factor" can be generalized as a "kernel" $K(x,t)=e^{-xt}$ (not the same as a kernel in linear algebra) and the general form is $\hat f(t)=\int_a^b K(x,t)f(x)dx$. For example $K(x,t)=e^{-ixt}$ gives the Fourier transform. Commented Aug 30, 2017 at 7:03
• the latter @Mark Commented Aug 30, 2017 at 7:14
• yes, a kernel, the "convergence factor" is much more like the kernel in integral transform @JeanMarie Commented Aug 30, 2017 at 7:19

For $x>0$, $\displaystyle \int_0^1 t^{x-1}dt=\frac{1}{x}$
• Instructive! But it doesn't have to be $\frac{1}{x}$, any other "convergence factor" with an integral is acceptable. Commented Aug 30, 2017 at 14:37