0
\begin{align}
\left[\frac{-2 b^2 s^2 \log \left(\frac{b \left(a+s^2\right)}{a}\right)-2 a b^2 \log \left(\frac{b \left(a+s^2\right)}{a}\right)
+2 b^2 s^2 \log \left(a+s^2\right)+2 a b^2 \log \left(a+s^2\right)
+2 b s^2 e^{\frac{b \left(a+s^2\right)}{a}} \text{Ei}\left(-\frac{b \left(s^2+a\right)}{a}\right)+2 b s^2 e^{\frac{b \left(a+s^2\right)}{a}} \log \left(\frac{b \left(a+s^2\right)}{a}\right)-2 b s^2 \log \left(\frac{b \left(a+s^2\right)}{a}\right)-2 a b \log \left(\frac{b \left(a+s^2\right)}{a}\right)-b s^2 e^{\frac{b \left(a+s^2\right)}{a}} \log \left(-\frac{b \left(a+s^2\right)}{a}\right)-2 (b+1) b \left(a+s^2\right) \log \left(\frac{a}{b}\right)+2 b s^2 \log \left(a+s^2\right)+2 a b \log \left(a+s^2\right)+b s^2 e^{\frac{b \left(a+s^2\right)}{a}} \log \left(-\frac{a}{b \left(a+s^2\right)}\right)+2 a}{2 a p},\Re\left(a+s^2\right)>0\land \left(\left(\Re\left(\frac{b}{a}\right)\geq 0\land \frac{b}{a}\neq 0\right)\lor \frac{b}{a}\notin \mathbb{R}\right)\right]
\end{align}
  • Hi, welcome. Start by removing the $. As align starts math mode on its own, you don't need/want those. – Torbjørn T. Aug 18 '17 at 7:04
  • Even after removing the outer pair of $ symbols, the code doesn't compile. Please fix any and all errors, so that readers of your posting will have a better shot at figuring out where to insert line breaks. – Mico Aug 18 '17 at 7:09
  • in the third line, \frac{... is wrong. – somkiat_t Aug 18 '17 at 7:19
  • the code compiles perfectly now.Please help me for splitting equation. – Nagesh Aug 18 '17 at 7:26
  • It's hard to know where to break the equation without knowing the line width, which depends on the documentclass and/or paper size. Also, I suggest you put some effort into making your LaTeX code readable to humans by inserting line breaks and indenting code, keeping more closely related parts together. The code is near impossible to navigate as is. – Harald Hanche-Olsen Aug 18 '17 at 7:33
0

Some Edition for you,

\begin{align}
\int_{\alpha=0}^{\infty}\lambda_1\lambda_2&\exp(-\lambda_1z)\frac{1} {\lambda_1 z+b}\frac{\exp(-\alpha h_{T_x,1}^2)}{p}d\alpha\nonumber\\
=&
\left.-2 b^2 s^2 \log \left(\frac{b \left(a+s^2\right)}{a}\right)-2 a b^2 \log \left(\frac{b \left(a+s^2\right)}{a}\right)+2 b^2 s^2\right.\nonumber\\
&\times \log \left(a+s^2\right)+2 a b^2 \log \left(a+s^2\right)+2 b s^2 e^{\frac{b \left(a+s^2\right)}{a}} \text{Ei}\left(-\frac{b \left(s^2+a\right)}{a}\right)\nonumber\\
&+2 b s^2 e^{\frac{b \left(a+s^2\right)}{a}} \log \left(\frac{b \left(a+s^2\right)}{a}\right)-2 b s^2 \log \left(\frac{b \left(a+s^2\right)}{a}\right)-2 a b \log \left(\frac{b \left(a+s^2\right)}{a}\right)\nonumber\\
&-b s^2 e^{\frac{b \left(a+s^2\right)}{a}} \log \left(-\frac{b \left(a+s^2\right)}{a}\right)-2 (b+1) b \left(a+s^2\right) \log \left(\frac{a}{b}\right)+2 b s^2 \log \left(a+s^2\right)\nonumber\\
&+2 a b \log \left(a+s^2\right)+b s^2 e^{\frac{b \left(a+s^2\right)}{a}} \log \left(-\frac{a}{b \left(a+s^2\right)}\right)+2 a{2 a p},\Re\left(a+s^2\right)>0\nonumber\\
&\land \left(\left(\Re\left(\frac{b}{a}\right)\geq 0
\land \frac{b}{a}\neq 0\right)\lor \frac{b}{a}\notin \mathbb{R}\right)
\end{align}
| improve this answer | |
  • 2
    You should use &=, not =&, in alignments. That way, spacing comes out right, as the environment is designed. Also, I suggest replacing &+ with &\phantom{{}=}+$, and similarly with other lines starting with a binary operation following the ampersand. That puts in an invisible equals sign for you. The {}` before the equals sign is to make TeX put in the space that ordinarily comes before a relation symbol (the same way align does it behind the curtains). – Harald Hanche-Olsen Aug 18 '17 at 7:38
1

I suggest avoiding the two-story fractions as long as possible, particularly in exponents.

Also, the denominator can be placed at the start and the conditions can be added on a line by themselves.

\documentclass{article}
\usepackage{amsmath,amssymb}

\DeclareMathOperator{\Ei}{Ei}

\begin{document}

\begin{align*}
\frac{1}{2ap}\biggl(
&
-2 b^2 s^2 \log(b(a+s^2)/a)
-2 a b^2 \log(b (a+s^2)/a)
+2 b^2 s^2 \log(a+s^2)
\\
&
+2 a b^2 \log (a+s^2)
+2 b s^2 e^{b(a+s^2)/a} \Ei(-b(s^2+a)/a)
\\
&
+2 b s^2 e^{b(a+s^2)/a} \log(b(a+s^2)/a)
-2 b s^2 \log(b(a+s^2)/a)
\\
&
-2 a b \log(b(a+s^2)/a)
-b s^2 e^{b(a+s^2)/a} \log (-b(a+s^2)/a)
\\
&
-2 (b+1) b(a+s^2) \log(a/b)
+2 b s^2 \log(a+s^2)
+2 a b \log(a+s^2)
\\
&
+b s^2 e^{b(a+s^2)/a} \log\Bigl(-\frac{a}{b(a+s^2)}\Bigr)
+2 a
\biggr)
\\[2ex]
&\Re\left(a+s^2\right)>0 \land 
\left(\left(\Re\left(\frac{b}{a}\right)\geq 0 \land
  \frac{b}{a}\neq 0\right)\lor \frac{b}{a}\notin \mathbb{R}\right)
\end{align*}

\end{document}

enter image description here

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