Consider the divergent integral ∫ − 1 1 1 t 2 d t = ( lim a → 0 − ∫ − 1 a 1 t 2 d t ) + ( lim b → 0 + ∫ b 1 1 t 2 d t ) = lim a → 0 − ( − 1 a − 1 ) + lim b → 0 + ( − 1 + 1 b ) = + ∞ {\displaystyle \int _{-1}^{1}{\frac {1}{t^{2}}}\,dt=\left(\lim _{a\to 0^{-}}\int _{-1}^{a}{\frac {1}{t^{2}}}\,dt\right)+\left(\lim _{b\to 0^{+}}\int _{b}^{1}{\frac {1}{t^{2}}}\,dt\right)=\lim _{a\to 0^{-}}\left(-{\frac {1}{a}}-1\right)+\lim _{b\to 0^{+}}\left(-1+{\frac {1}{b}}\right)=+\infty } Its Cauchy principal value also diverges since C ∫ − 1 1 1 t 2 d t = lim ε → 0 + ( ∫ − 1 − ε 1 t 2 d t + ∫ ε 1 1 t 2 d t ) = lim ε → 0 + ( 1 ε − 1 − 1 + 1 ε ) = + ∞ {\displaystyle {\mathcal {C}}\int _{-1}^{1}{\frac {1}{t^{2}}}\,dt=\lim _{\varepsilon \to 0^{+}}\left(\int _{-1}^{-\varepsilon }{\frac {1}{t^{2}}}\,dt+\int _{\varepsilon }^{1}{\frac {1}{t^{2}}}\,dt\right)=\lim _{\varepsilon \to 0^{+}}\left({\frac {1}{\varepsilon }}-1-1+{\frac {1}{\varepsilon }}\right)=+\infty } To assign a finite value to this divergent integral, we may consider H ∫ − 1 1 1 t 2 d t = H ∫ − 1 1 1 ( t − x ) 2 d t | x = 0 = d d x ( C ∫ − 1 1 1 t − x d t ) | x = 0 {\displaystyle {\mathcal {H}}\int _{-1}^{1}{\frac {1}{t^{2}}}\,dt={\mathcal {H}}\int _{-1}^{1}{\frac {1}{(t-x)^{2}}}\,dt{\Bigg |}_{x=0}={\frac {d}{dx}}\left({\mathcal {C}}\int _{-1}^{1}{\frac {1}{t-x}}\,dt\right){\Bigg |}_{x=0}} The inner Cauchy principal value is given by C ∫ − 1 1 1 t − x d t = lim ε → 0 + ( ∫ − 1 − ε 1 t − x d t + ∫ ε 1 1 t − x d t ) = lim ε → 0 + ( ln | ε + x 1 + x | + ln | 1 − x ε − x | ) = ln | 1 − x 1 + x | {\displaystyle {\mathcal {C}}\int _{-1}^{1}{\frac {1}{t-x}}\,dt=\lim _{\varepsilon \to 0^{+}}\left(\int _{-1}^{-\varepsilon }{\frac {1}{t-x}}\,dt+\int _{\varepsilon }^{1}{\frac {1}{t-x}}\,dt\right)=\lim _{\varepsilon \to 0^{+}}\left(\ln \left|{\frac {\varepsilon +x}{1+x}}\right|+\ln \left|{\frac {1-x}{\varepsilon -x}}\right|\right)=\ln \left|{\frac {1-x}{1+x}}\right|} Therefore, H ∫ − 1 1 1 t 2 d t = d d x ( ln | 1 − x 1 + x | ) | x = 0 = 2 x 2 − 1 | x = 0 = − 2 {\displaystyle {\mathcal {H}}\int _{-1}^{1}{\frac {1}{t^{2}}}\,dt={\frac {d}{dx}}\left(\ln \left|{\frac {1-x}{1+x}}\right|\right){\Bigg |}_{x=0}={\frac {2}{x^{2}-1}}{\Bigg |}_{x=0}=-2} Note that this value does not represent the area under the curve y(t) = 1/t2, which is clearly always positive. However, it can be seen where this comes from. Recall the Cauchy principal value of this integral, when evaluated at the endpoints, took the form lim ε → 0 + ( 1 ε − 1 − 1 + 1 ε ) = + ∞ {\displaystyle \lim _{\varepsilon \to 0^{+}}\left({\frac {1}{\varepsilon }}-1-1+{\frac {1}{\varepsilon }}\right)=+\infty }
If one removes the infinite components, the pair of 1 ε {\displaystyle {\frac {1}{\varepsilon }}} terms, that which remains is lim ε → 0 + ( − 1 − 1 ) = − 2 {\displaystyle \lim _{\varepsilon \to 0^{+}}\left(-1-1\right)=-2}
which equals the value derived above.