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<div class="titlepage"><div><div><h3 class="title">
<a name="math_toolkit.double_exponential.de_caveats"></a><a class="link" href="de_caveats.html" title="Caveats">Caveats</a>
</h3></div></div></div>
<p>
        A few things to keep in mind while using the tanh-sinh, exp-sinh, and sinh-sinh
        quadratures:
      </p>
<p>
        These routines are <span class="bold"><strong>very</strong></span> aggressive about
        approaching the endpoint singularities. This allows lots of significant digits
        to be extracted, but also has another problem: Roundoff error can cause the
        function to be evaluated at the endpoints. A few ways to avoid this: Narrow
        up the bounds of integration to say, [a + &#949;, b - &#949;], make sure (a+b)/2 and
        (b-a)/2 are representable, and finally, if you think the compromise between
        accuracy an usability has gone too far in the direction of accuracy, file
        a ticket.
      </p>
<p>
        Both exp-sinh and sinh-sinh quadratures evaluate the functions they are passed
        at <span class="bold"><strong>very</strong></span> large argument. You might understand
        that x<sup>12</sup>exp(-x) is should be zero when x<sup>12</sup> overflows, but IEEE floating point
        arithmetic does not. Hence <code class="computeroutput"><span class="identifier">std</span><span class="special">::</span><span class="identifier">pow</span><span class="special">(</span><span class="identifier">x</span><span class="special">,</span> <span class="number">12</span><span class="special">)*</span><span class="identifier">std</span><span class="special">::</span><span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)</span></code> is an indeterminate form whenever <code class="computeroutput"><span class="identifier">std</span><span class="special">::</span><span class="identifier">pow</span><span class="special">(</span><span class="identifier">x</span><span class="special">,</span>
        <span class="number">12</span><span class="special">)</span></code>
        overflows. So make sure your functions have the correct limiting behavior;
        for example
      </p>
<pre class="programlisting"><span class="keyword">auto</span> <span class="identifier">f</span> <span class="special">=</span> <span class="special">[](</span><span class="keyword">double</span> <span class="identifier">x</span><span class="special">)</span> <span class="special">{</span>
    <span class="keyword">double</span> <span class="identifier">t</span> <span class="special">=</span> <span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">);</span>
    <span class="keyword">if</span><span class="special">(</span><span class="identifier">t</span> <span class="special">==</span> <span class="number">0</span><span class="special">)</span>
    <span class="special">{</span>
        <span class="keyword">return</span> <span class="number">0</span><span class="special">;</span>
    <span class="special">}</span>
    <span class="keyword">return</span> <span class="identifier">t</span><span class="special">*</span><span class="identifier">pow</span><span class="special">(</span><span class="identifier">x</span><span class="special">,</span> <span class="number">12</span><span class="special">);</span>
<span class="special">};</span>
</pre>
<p>
        has the correct behavior for large <span class="emphasis"><em>x</em></span>, but <code class="computeroutput"><span class="keyword">auto</span> <span class="identifier">f</span> <span class="special">=</span> <span class="special">[](</span><span class="keyword">double</span>
        <span class="identifier">x</span><span class="special">)</span> <span class="special">{</span> <span class="keyword">return</span> <span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)*</span><span class="identifier">pow</span><span class="special">(</span><span class="identifier">x</span><span class="special">,</span> <span class="number">12</span><span class="special">);</span> <span class="special">};</span></code> does
        not.
      </p>
<p>
        Oscillatory integrals, such as the sinc integral, are poorly approximated
        by double-exponential quadrature. Fortunately the error estimates and L1
        norm are massive for these integrals, but nonetheless, oscillatory integrals
        require different techniques.
      </p>
<p>
        A special mention should be made about integrating through zero: while our
        range adaptors preserve precision when one endpoint is zero, things get harder
        when the origin is neither in the center of the range, nor at an endpoint.
        Consider integrating:
      </p>
<div class="blockquote"><blockquote class="blockquote"><p>
          <span class="serif_italic">1 / (1 +x^2)</span>
        </p></blockquote></div>
<p>
        Over (a, &#8734;). As long as <code class="computeroutput"><span class="identifier">a</span> <span class="special">&gt;=</span> <span class="number">0</span></code> both
        the tanh_sinh and the exp_sinh integrators will handle this just fine: in
        fact they provide a rather efficient method for this kind of integral. However,
        if we have <code class="computeroutput"><span class="identifier">a</span> <span class="special">&lt;</span>
        <span class="number">0</span></code> then we are forced to adapt the range
        in a way that produces abscissa values near zero that have an absolute error
        of &#949;, and since all of the area of the integral is near zero both integrators
        thrash around trying to reach the target accuracy, but never actually get
        there for <code class="computeroutput"><span class="identifier">a</span> <span class="special">&lt;&lt;</span>
        <span class="number">0</span></code>. On the other hand, the simple expedient
        of breaking the integral into two domains: (a, 0) and (0, b) and integrating
        each seperately using the tanh-sinh integrator, works just fine.
      </p>
<p>
        Finally, some endpoint singularities are too strong to be handled by <code class="computeroutput"><span class="identifier">tanh_sinh</span></code> or equivalent methods, for example
        consider integrating the function:
      </p>
<pre class="programlisting"><span class="keyword">double</span> <span class="identifier">p</span> <span class="special">=</span> <span class="identifier">some_value</span><span class="special">;</span>
<span class="identifier">tanh_sinh</span><span class="special">&lt;</span><span class="keyword">double</span><span class="special">&gt;</span> <span class="identifier">integrator</span><span class="special">;</span>
<span class="keyword">auto</span> <span class="identifier">f</span> <span class="special">=</span> <span class="special">[&amp;](</span><span class="keyword">double</span> <span class="identifier">x</span><span class="special">){</span> <span class="keyword">return</span> <span class="identifier">pow</span><span class="special">(</span><span class="identifier">tan</span><span class="special">(</span><span class="identifier">x</span><span class="special">),</span> <span class="identifier">p</span><span class="special">);</span> <span class="special">};</span>
<span class="keyword">auto</span> <span class="identifier">Q</span> <span class="special">=</span> <span class="identifier">integrator</span><span class="special">.</span><span class="identifier">integrate</span><span class="special">(</span><span class="identifier">f</span><span class="special">,</span> <span class="number">0</span><span class="special">,</span> <span class="identifier">constants</span><span class="special">::</span><span class="identifier">half_pi</span><span class="special">&lt;</span><span class="keyword">double</span><span class="special">&gt;());</span>
</pre>
<p>
        The first problem with this function, is that the singularity is at &#960;/2, so
        if we're integrating over (0, &#960;/2) then we can never approach closer to the
        singularity than &#949;, and for p less than but close to 1, we need to get <span class="emphasis"><em>very</em></span>
        close to the singularity to find all the area under the function. If we recall
        the identity <code class="literal">tan(&#960;/2 - x) == 1/tan(x)</code> then we can rewrite
        the function like this:
      </p>
<pre class="programlisting"><span class="keyword">auto</span> <span class="identifier">f</span> <span class="special">=</span> <span class="special">[&amp;](</span><span class="keyword">double</span> <span class="identifier">x</span><span class="special">){</span> <span class="keyword">return</span> <span class="identifier">pow</span><span class="special">(</span><span class="identifier">tan</span><span class="special">(</span><span class="identifier">x</span><span class="special">),</span> <span class="special">-</span><span class="identifier">p</span><span class="special">);</span> <span class="special">};</span>
</pre>
<p>
        And now the singularity is at the origin and we can get much closer to it
        when evaluating the integral: all we have done is swap the integral endpoints
        over.
      </p>
<p>
        This actually works just fine for p &lt; 0.95, but after that the <code class="computeroutput"><span class="identifier">tanh_sinh</span></code> integrator starts thrashing around
        and is unable to converge on the integral. The problem is actually a lack
        of exponent range: if we simply swap type double for something with a greater
        exponent range (an 80-bit long double or a quad precision type), then we
        can get to at least p = 0.99. If we want to go beyond that, or stick with
        type double, then we have to get smart.
      </p>
<p>
        The easiest method is to notice that for small x, then <code class="literal">tan(x) &#8773; x</code>,
        and so we are simply integrating x<sup>-p</sup>. Therefore we can use this approximation
        over (0, small), and integrate numerically from (small, &#960;/2), using &#949; as a suitable
        crossover point seems sensible:
      </p>
<pre class="programlisting"><span class="keyword">double</span> <span class="identifier">p</span> <span class="special">=</span> <span class="identifier">some_value</span><span class="special">;</span>
<span class="keyword">double</span> <span class="identifier">crossover</span> <span class="special">=</span> <span class="identifier">std</span><span class="special">::</span><span class="identifier">numeric_limits</span><span class="special">&lt;</span><span class="keyword">double</span><span class="special">&gt;::</span><span class="identifier">epsilon</span><span class="special">();</span>
<span class="identifier">tanh_sinh</span><span class="special">&lt;</span><span class="keyword">double</span><span class="special">&gt;</span> <span class="identifier">integrator</span><span class="special">;</span>
<span class="keyword">auto</span> <span class="identifier">f</span> <span class="special">=</span> <span class="special">[&amp;](</span><span class="keyword">double</span> <span class="identifier">x</span><span class="special">){</span> <span class="keyword">return</span> <span class="identifier">pow</span><span class="special">(</span><span class="identifier">tan</span><span class="special">(</span><span class="identifier">x</span><span class="special">),</span> <span class="identifier">p</span><span class="special">);</span> <span class="special">};</span>
<span class="keyword">auto</span> <span class="identifier">Q</span> <span class="special">=</span> <span class="identifier">integrator</span><span class="special">.</span><span class="identifier">integrate</span><span class="special">(</span><span class="identifier">f</span><span class="special">,</span> <span class="identifier">crossover</span><span class="special">,</span> <span class="identifier">constants</span><span class="special">::</span><span class="identifier">half_pi</span><span class="special">&lt;</span><span class="keyword">double</span><span class="special">&gt;())</span> <span class="special">+</span> <span class="identifier">pow</span><span class="special">(</span><span class="identifier">crossover</span><span class="special">,</span> <span class="number">1</span> <span class="special">-</span> <span class="identifier">p</span><span class="special">)</span> <span class="special">/</span> <span class="special">(</span><span class="number">1</span> <span class="special">-</span> <span class="identifier">p</span><span class="special">);</span>
</pre>
<p>
        There is an alternative, more complex method, which is applicable when we
        are dealing with expressions which can be simplified by evaluating by logs.
        Let's suppose that as in this case, all the area under the graph is infinitely
        close to zero, now inagine that we could expand that region out over a much
        larger range of abscissa values: that's exactly what happens if we perform
        argument substitution, replacing <code class="computeroutput"><span class="identifier">x</span></code>
        by <code class="computeroutput"><span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)</span></code> (note
        that we must also multiply by the derivative of <code class="computeroutput"><span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)</span></code>).
        Now the singularity at zero is moved to +&#8734;, and the &#960;/2 bound to -log(&#960;/2).
        Initially our argument substituted function looks like:
      </p>
<pre class="programlisting"><span class="keyword">auto</span> <span class="identifier">f</span> <span class="special">=</span> <span class="special">[&amp;](</span><span class="keyword">double</span> <span class="identifier">x</span><span class="special">){</span> <span class="keyword">return</span> <span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)</span> <span class="special">*</span> <span class="identifier">pow</span><span class="special">(</span><span class="identifier">tan</span><span class="special">(</span><span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)),</span> <span class="special">-</span><span class="identifier">p</span><span class="special">);</span> <span class="special">};</span>
</pre>
<p>
        Which is hardly any better, as we still run out of exponent range just as
        before. However, if we replace <code class="computeroutput"><span class="identifier">tan</span><span class="special">(</span><span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">))</span></code> by
        <code class="computeroutput"><span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)</span></code> for
        suitably small <code class="computeroutput"><span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)</span></code>, and
        therefore <code class="literal">x &gt; -log(&#949;)</code>, we can greatly simplify the expression
        and evaluate by logs:
      </p>
<pre class="programlisting"><span class="keyword">auto</span> <span class="identifier">f</span> <span class="special">=</span> <span class="special">[&amp;](</span><span class="keyword">double</span> <span class="identifier">x</span><span class="special">)</span>
<span class="special">{</span>
   <span class="keyword">static</span> <span class="keyword">const</span> <span class="keyword">double</span> <span class="identifier">crossover</span> <span class="special">=</span> <span class="special">-</span><span class="identifier">log</span><span class="special">(</span><span class="identifier">std</span><span class="special">::</span><span class="identifier">numeric_limits</span><span class="special">&lt;</span><span class="keyword">double</span><span class="special">&gt;::</span><span class="identifier">epsilon</span><span class="special">());</span>
   <span class="keyword">return</span> <span class="identifier">x</span> <span class="special">&gt;</span> <span class="identifier">crossover</span> <span class="special">?</span> <span class="identifier">exp</span><span class="special">((</span><span class="identifier">p</span> <span class="special">-</span> <span class="number">1</span><span class="special">)</span> <span class="special">*</span> <span class="identifier">x</span><span class="special">)</span> <span class="special">:</span> <span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)</span> <span class="special">*</span> <span class="identifier">pow</span><span class="special">(</span><span class="identifier">tan</span><span class="special">(</span><span class="identifier">exp</span><span class="special">(-</span><span class="identifier">x</span><span class="special">)),</span> <span class="special">-</span><span class="identifier">p</span><span class="special">);</span>
<span class="special">};</span>
</pre>
<p>
        This form integrates just fine over (-log(&#960;/2), +&#8734;) using either the <code class="computeroutput"><span class="identifier">tanh_sinh</span></code> or <code class="computeroutput"><span class="identifier">exp_sinh</span></code>
        classes.
      </p>
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