# Integral evalutation

I am trying to integrate the following expression over $L$.

E^(-2 L n) (1 - L/(2 s))^(-1 + 4 n µ) (L/s)^(-1 +  4 n v)


I did...

Integrate[E^(-2 L n) (1 - L/(2 s))^(-1 + 4 n µ) (L/s)^(-1 +  4 n v), L]


but Mathematica fails to solve this integral. I tried with actual values but this issue remains.

Integrate[ E^(-2 L 100) (1 - L/(2 0.08))^(-1 + 4 100 3*^-7) (L/0.08)^(-1 + 4 100 3*^-7), L]


How can I calculate or approximate this integral?

Note the following assumptions

0 <= s <= 1
0 <= µ <= 1
0 <= v <= 1
1 <= n <= Infinity


, where the sign <= means "smaller or equal". Note also that µ and v are typically very small (on the order of $10^{-7}$)

Setting

integrand= E^(-2 L 100) (1 - L/(2 0.08))^(-1 + 4 100 3*^-7) (L/0.08)^(-1 + 4 100 3*^-7)


with the example given in your question, you can use

ni[x_]:=NIntegrate[integrand,{L,0.1,x}]


and

Plot[Re@ni[x],{x,0.1,0.5}]


to plot the numerical integrand. Your example does not seem to lend itself well to approximation, though. Or perhabs I chose the L in the wrong interval? Also, there may not exist a "nice" solution to your integral.

• Thanks for your answer @Berg. The range of values for L is slightly high but correspond to what I am interested in. Can you point me to some source of information so that I can understand the expression Re@ that might explain why the plot looks like a cardiogram rather than like a continuously increasing function of $x$! – Remi.b Oct 27 '14 at 23:48
• Not sure about a reference, but numerical approximations can fail, if, for example, the numerical values involved are very small. In that case the floating point reals in the computer cannot properly store those values. One could try to rescale the problem. Another problem can appear if one has highly oscillating integrals. Think of a, say, a sinus function with very high frequency. One then has lots of parts that cancel, but the numerical approximation only evaluates some of those. Depending on what points exactly are used for the approximation the resulting integral can be large or small. – Berg Oct 28 '14 at 8:43
• Oh, and Re@x=Re[x] is just the real part of x. – Berg Oct 28 '14 at 8:53