Improper integral with an … Improper integrals are said to be convergent if the limit is ﬁnite and that limit is the value of the improper integral. In other words, indefinite integration is an opposite operation to analytical differentiation. Otherwise the improper integrals are divergent. Solution. Wolfram|Alpha is a great tool for calculating antiderivatives and definite integrals, double and triple integrals, and improper integrals. $\int_0^{\infty}\left(\frac{1}{1+x^2}\right)dx$, $\left[\frac{1}{\sqrt{1}}\arctan\left(\frac{x}{\sqrt{1}}\right)\right]_{0}^{\infty }$, $\left[\frac{1}{1}\arctan\left(\frac{x}{\sqrt{1}}\right)\right]_{0}^{\infty }$, $\left[\arctan\left(x\right)\right]_{0}^{\infty }$, $\lim_{c\to\infty }\left(\left[\arctan\left(x\right)\right]_{0}^{c}\right)$, $\lim_{c\to\infty }\left(\arctan\left(c\right)-\arctan\left(0\right)\right)$, $\lim_{c\to\infty }\left(\arctan\left(c\right)-1\cdot 0\right)$, $\lim_{c\to\infty }\left(\arctan\left(c\right)\right)$. Be patient! Improper Integral Calculator is a free online tool that displays the integrated value for the improper integral. This equation reads as follows: the indefinite integral of with respect to .A definite integral has limits of integration. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Let {f\left( x \right)}f(x) be a continuous function on the interval \left[ {a,\infty} \right). 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