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I need to compute the definite integral.

$$\int_{-k}^{k} \frac{e^{-ax-bx^2}}{\sqrt{1-x^2}}dx$$

Although the integral looks neat and fairly simple in form, I'm not able to evaluate it further. I tried evaluating it using integration by parts but without success. I also couldn't find the solution in the book, "Table of Integrals, Series, and Products" by Gradshteyn and Ryzhik.

Can anyone help me in solving this?

Thank you.

Spandan Kukade
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JayanthJ
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    I couldn't prove it to you, but I'm almost positive this doesn't have a closed form. Take $(k,a,b) = (\frac{1}{2},1,0)$, and MatLab is unable to give an answer. When you mess with these $\sqrt{1-x^2}$ and transcendental functions, it's rare you get a closed form. Would love to be proved wrong, though! – Moni145 Jul 16 '21 at 10:39
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    Basically this is indefinite integral (for $k$ being arbitrary). – metamorphy Jul 16 '21 at 12:17

1 Answers1

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You never specified a closed form in your question, so here is an exact form answer using the following: integration source, hypergeometric argument simplification, main simplification, signum or sign function, proof of series expansion, an expansion of exp(y), binomial theorem,incomplete beta function and hypergeometric function.

Using the Maclaurin series expansion for exp(y) and binomial theorem: $$\mathrm{\int_{-k}^k \frac{e^{-bx^2-ax}}{\sqrt{1-x^2}}dx= \int_{-k}^k \frac{1}{\sqrt{1-x^2}}\sum_{n=0}^\infty\frac{(-bx^2-ax)^n}{n!}dx= \int_{-k}^k \frac{1}{\sqrt{1-x^2}}\sum_{n=0}^\infty\frac{(-1)^n}{n!} \sum_{m=0}^n \binom nm b^ma^{n-m}x^{m+n}dx}$$

Integrating with the hypergeometric function seen above gives after a little simplification. The end is sped up with bound substitution already applied:

$$\mathrm{\int_{-k}^k \frac{e^{-bx^2-ax}}{\sqrt{1-x^2}}dx= \sum_{n=0}^\infty(-1)^n\sum_{m=0}^n \frac{b^ma^{n-m}}{m!(n-m)!} \int_{-k}^k \frac{x^{m+n}}{\sqrt{1-x^2}}dx=\quad\sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}k^{m+n+1}\left[(-1)^m+(-1)^n\right]\,_2F_1\left(\frac12,\frac{m+n+1}{2}, \frac{m+n+3}{2},k^2\right)}{m!(n-m)!(m+n+1)}= \quad \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}k^{m+n+1}\left[(-1)^m+(-1)^n\right]\left(\frac{m+n+1}{2|k|^{m+n+1}}B_{k^2}\left(\frac{m+n+1}{2},\frac12\right)\right)}{m!(n-m)!(m+n+1)}= \quad\frac12 \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}sgn\left(k^{m+n+1}\right)\left[(-1)^m+(-1)^n\right]B_{k^2}\left(\frac{m+n+1}{2},\frac12\right)}{m!(n-m)!}}$$

Here is graphical proof of this result. Notice the inverse sine look to the graph.

Feel free to simplify this as the consecutive $(-1)^r$ terms look like they can be simplified. Go ahead and try this result if you like. The “proof of expansion link” takes a bit of time to load, so please be patient. Please correct me and give me feedback!

Тyma Gaidash
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    That's quite an answer you got there. – Rounak Sarkar Jul 16 '21 at 14:11
  • This was really helpful. I will try to simplify the double summation in the last step even further. Thanks a lot! – JayanthJ Jul 17 '21 at 08:05
  • @JayanthJayakumar I have simplified the hypergeometric function. The alternating piece can be separated into two series. There may be a typo. What is this problem for? – Тyma Gaidash Jul 17 '21 at 15:18
  • @TymaGaidash, I tried simplifying the last step by first changing the order of summation using a double infinite sum identity since the double summation, as it stands, is difficult to evaluate analytically. Secondly, after separating into two series, I tried using some known identities for the Beta function to get a more compact form and then perform the said summation. However, I was not successful in both these attempts. – JayanthJ Jul 22 '21 at 14:24
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    This problem is a small part of a bigger physics research problem where I need to compute the Wigner function corresponding to a state of light. By the way, do you have any more ideas/suggestions to simplify the double summation, just in case? – JayanthJ Jul 22 '21 at 14:28
  • @JayanthJayakumar Please check the answer as there may he a typo, as I am human. You can split up the sum with the factored $[(-1)^m+(-1)^n]$. Here is a great example on a nice simplification:Post with double sum simplification. Tell me if this helps. – Тyma Gaidash Jul 22 '21 at 14:38
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    @TymaGaidash, I checked your answer by redoing it in a similar way and there was no typo. Further, I was able to split the sum as follows (from the last step). Please let me know if this correct. $= \sum_{s=0}^{\infty} \sum_{r=0}^{s} \frac{b^{2r} a^{2(s-r)}}{2r! (2(s-r))!} \text{sgn}(k^{2(r+s)+1}) B_{k^2} \big(\frac{2(r+s)+1}{2},\frac{1}{2}\big) - \sum_{s=0}^{\infty} \sum_{r=0}^{s} \frac{b^{2r+1} a^{2(s-r)}}{(2r+1)! (2(s-r))!} \text{sgn}(k^{2(r+s)+3}) B_{k^2} \big(\frac{2(r+s)+3}{2},\frac{1}{2}\big)$. – JayanthJ Jul 23 '21 at 15:55
  • @JayanthJayakumar Whoops, the graph should be the the square root part in the bottom. This actually looks correct, just do the (2r)! and not 2r!. You are correct! New graph. Notice how the integral and your result have the same graph. – Тyma Gaidash Jul 23 '21 at 18:31
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    @JayanthJayakumar Here are results for a simplified answer. – Тyma Gaidash Jul 24 '21 at 19:50
  • @RounakSarkar It can be simplified with a special function in this edited version. – Тyma Gaidash Oct 20 '21 at 20:57
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    @TymaGaidash. I love how you saw my comment like 2 months later. It's ok – Rounak Sarkar Oct 21 '21 at 07:18
  • @RounakSarkar I guess we met before. I also saw your $\int \text{Si}^n(x) dx$ post. – Тyma Gaidash Oct 21 '21 at 11:59
  • @TymaGaidash. I am not sure what do you mean by 'we have met before'. Yeah we have talked here in this site before. – Rounak Sarkar Oct 21 '21 at 12:47