Plot of the <a href="https://en.wikipedia.org/wiki/Inverse_tangent_integral" class="extiw" title="en:Inverse tangent integral">inverse tangent integrals</a> Ti<sub>0</sub>(x), Ti<sub>1</sub>(x), Ti<sub>2</sub>(x) and Ti<sub>3</sub>(x) in the interval [-5,5]:
<ul><li><span class="mwe-math-element"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \operatorname {Ti} _{0}(z)={z \over 1+z^{2}}}"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mi>Ti</mi><mrow class="MJX-TeXAtom-ORD"><mn>0</mn></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mi>z</mi><mrow><mn>1</mn><mo>+</mo><msup><mi>z</mi><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \operatorname {Ti} _{0}(z)={z \over 1+z^{2}}}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/41dfd98a0f80021f4ecbdfc9456904d03c8e1df6" class="mwe-math-fallback-image-inline" aria-hidden="true" style="vertical-align: -2.338ex; width:16.359ex; height:5.176ex;" alt="{\displaystyle \operatorname {Ti} _{0}(z)={z \over 1+z^{2}}}"></span></li>
<li><span class="mwe-math-element"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \operatorname {Ti} _{1}(z)=\arctan z}"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mi>Ti</mi><mrow class="MJX-TeXAtom-ORD"><mn>1</mn></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo>=</mo><mi>arctan</mi><mo>⁡<!-- ⁡ --></mo><mi>z</mi></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \operatorname {Ti} _{1}(z)=\arctan z}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/38de7fe56258cf0c8f6a50ea5f973667132469ee" class="mwe-math-fallback-image-inline" aria-hidden="true" style="vertical-align: -0.838ex; width:17.316ex; height:2.843ex;" alt="{\displaystyle \operatorname {Ti} _{1}(z)=\arctan z}"></span></li>
<li><span class="mwe-math-element"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \operatorname {Ti} _{2}(z)=\int _{0}^{z}{\arctan t \over t}\,\mathrm {d} t}"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mi>Ti</mi><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo>=</mo><msubsup><mo>∫<!-- ∫ --></mo><mrow class="MJX-TeXAtom-ORD"><mn>0</mn></mrow><mrow class="MJX-TeXAtom-ORD"><mi>z</mi></mrow></msubsup><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mi>arctan</mi><mo>⁡<!-- ⁡ --></mo><mi>t</mi></mrow><mi>t</mi></mfrac></mrow><mspace width="thinmathspace"></mspace><mrow class="MJX-TeXAtom-ORD"><mi mathvariant="normal">d</mi></mrow><mi>t</mi></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \operatorname {Ti} _{2}(z)=\int _{0}^{z}{\arctan t \over t}\,\mathrm {d} t}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/66e830610471d988418e3c6051ae742c318ffd1a" class="mwe-math-fallback-image-inline" aria-hidden="true" style="vertical-align: -2.338ex; width:24.276ex; height:5.843ex;" alt="{\displaystyle \operatorname {Ti} _{2}(z)=\int _{0}^{z}{\arctan t \over t}\,\mathrm {d} t}"></span></li>
<li><span class="mwe-math-element"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \operatorname {Ti} _{3}(z)={1 \over 2i}\left[\operatorname {Li} _{3}(iz)-\operatorname {Li} _{3}(-iz)\right]}"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mi>Ti</mi><mrow class="MJX-TeXAtom-ORD"><mn>3</mn></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mn>1</mn><mrow><mn>2</mn><mi>i</mi></mrow></mfrac></mrow><mrow><mo>[</mo><mrow><msub><mi>Li</mi><mrow class="MJX-TeXAtom-ORD"><mn>3</mn></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mi>i</mi><mi>z</mi><mo stretchy="false">)</mo><mo>−<!-- − --></mo><msub><mi>Li</mi><mrow class="MJX-TeXAtom-ORD"><mn>3</mn></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mo>−<!-- − --></mo><mi>i</mi><mi>z</mi><mo stretchy="false">)</mo></mrow><mo>]</mo></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \operatorname {Ti} _{3}(z)={1 \over 2i}\left[\operatorname {Li} _{3}(iz)-\operatorname {Li} _{3}(-iz)\right]}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9bfe101d51307b339921187b299a0cc7fca9d7f3" class="mwe-math-fallback-image-inline" aria-hidden="true" style="vertical-align: -1.838ex; width:32.213ex; height:5.176ex;" alt="{\displaystyle \operatorname {Ti} _{3}(z)={1 \over 2i}\left[\operatorname {Li} _{3}(iz)-\operatorname {Li} _{3}(-iz)\right]}"></span></li>
<li><span class="mwe-math-element"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \operatorname {Ti} _{n+1}(z)=\int _{0}^{z}{\operatorname {Ti} _{n}(t) \over t}\,\mathrm {d} t}"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mi>Ti</mi><mrow class="MJX-TeXAtom-ORD"><mi>n</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo>=</mo><msubsup><mo>∫<!-- ∫ --></mo><mrow class="MJX-TeXAtom-ORD"><mn>0</mn></mrow><mrow class="MJX-TeXAtom-ORD"><mi>z</mi></mrow></msubsup><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><msub><mi>Ti</mi><mrow class="MJX-TeXAtom-ORD"><mi>n</mi></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo></mrow><mi>t</mi></mfrac></mrow><mspace width="thinmathspace"></mspace><mrow class="MJX-TeXAtom-ORD"><mi mathvariant="normal">d</mi></mrow><mi>t</mi></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \operatorname {Ti} _{n+1}(z)=\int _{0}^{z}{\operatorname {Ti} _{n}(t) \over t}\,\mathrm {d} t}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/2f37fa7a3a1faddd126c8ba094b49eb6bc2fd9fd" class="mwe-math-fallback-image-inline" aria-hidden="true" style="vertical-align: -2.338ex; width:25.04ex; height:6.176ex;" alt="{\displaystyle \operatorname {Ti} _{n+1}(z)=\int _{0}^{z}{\operatorname {Ti} _{n}(t) \over t}\,\mathrm {d} t}"></span></li></ul><dl><dd><span class="mwe-math-element"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \operatorname {Ti} _{s}(z)={1 \over 2i}\left[\operatorname {Li} _{s}(iz)-\operatorname {Li} _{s}(-iz)\right]}"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mi>Ti</mi><mrow class="MJX-TeXAtom-ORD"><mi>s</mi></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mn>1</mn><mrow><mn>2</mn><mi>i</mi></mrow></mfrac></mrow><mrow><mo>[</mo><mrow><msub><mi>Li</mi><mrow class="MJX-TeXAtom-ORD"><mi>s</mi></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mi>i</mi><mi>z</mi><mo stretchy="false">)</mo><mo>−<!-- − --></mo><msub><mi>Li</mi><mrow class="MJX-TeXAtom-ORD"><mi>s</mi></mrow></msub><mo>⁡<!-- ⁡ --></mo><mo stretchy="false">(</mo><mo>−<!-- − --></mo><mi>i</mi><mi>z</mi><mo stretchy="false">)</mo></mrow><mo>]</mo></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \operatorname {Ti} _{s}(z)={1 \over 2i}\left[\operatorname {Li} _{s}(iz)-\operatorname {Li} _{s}(-iz)\right]}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/8be626e98592386822050d0b801d2eb2eea160ae" class="mwe-math-fallback-image-inline" aria-hidden="true" style="vertical-align: -1.838ex; width:32.06ex; height:5.176ex;" alt="{\displaystyle \operatorname {Ti} _{s}(z)={1 \over 2i}\left[\operatorname {Li} _{s}(iz)-\operatorname {Li} _{s}(-iz)\right]}"></span></dd></dl>
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