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This article shows a technique of evaluating a definite integral by introducing a suitable parameter. This however doesn't throw light on motivation for introducing that particular parameter.

For inctance:

$1)$ $\int_0^\infty \dfrac{\sin x}{x}dx $ can be evaluated by introducing the function $f(b)=\int_0^\infty \dfrac{\sin x}{x}e^{bx}dx$

$2)$ $\int_0^{\pi/2} x\cot x dx$ can be evaluated by introducing $f(b)=\int_0^{\pi/2}\dfrac{\arctan(b\tan(x))}{\tan (x)} dx$

What is the motivation behind these parameters, in general how would I find a parameter to evaluate a particular definite integral without trail and error?

Mathslover
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    I've never seen an exposition presenting a methodical approach to this method. It seems to rely on one's having a particularly good knack for guessing. It would be nice to see an exploration of how to evaluate some particular family of integrals at least, wouldn't it? – MPW Sep 16 '14 at 11:33
  • See https://math.stackexchange.com/a/3453393/686284 – Quanto Jul 18 '23 at 01:46

1 Answers1

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A general method does not exist; otherwise we could calculate all integrals, which is evidently not true.

We can say something about certain special cases though:

  1. If we have an integrand of the form $\dfrac{f(x)}{x}$, where $f$ is a simple function of $x$, it is often convenient to introduce a parameter such that upon differentiation, the $x$ in the denominator is removed. An application of this rule allows the evaluation of, for example, $$\int_0^{\infty} dx \, \dfrac{e^{-x} - e^{-2x}}{x}.$$
  2. This seems more like a lucky guess.

Another useful special case: for $\int dx \log(1+x) R(x)$ or $\int dx \tan^{-1}(x) R(x)$ with $R$ a rational function of $x$, put a parameter inside the $\log$ or $\arctan$. Upon differentiation, you will get a rational function which you can always integrate in principle.

user111187
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